Mathematics Tricks & Tips Of Probability For NDA

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Tricks & Tips Of Probability For NDA

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Probability Of an event

Q 2713191949

If two fair dice are thrown, then what is the probability that the sum is neither 8 nor 9?
NDA Paper 1 2017

(A)

`1/6`

(B)

`1/4`

(C)

`3/4`

(D)

`5/6`

Solution:

possible outcome for 8 `= {(2,6), (6,2), (3,5) , (5, 3),(4,4) }`

Possible outcome for 9 = `{(3,6), (6,3),(4,5),(5,4)}`

Probability of sum is 8 or 9 `= 9/36`

probability of sum is neither 8 nor 9 `= 1-9/36 = 3/4`

Correct Answer is `=>` (C) `3/4`

Q 2147280183

A card is drawn from a well-shuffled deck of `52` cards.
What is the probability that it is queen of spade?
NDA Paper 1 2016

(A)

`1 /(52)`

(B)

`1 /(13)`

(C)

`1 /4`

(D)

`1 /8`

Solution:

A card can be chosen from a pack of `52` cards in `text()^(52)C_(1)`
ways = `52` ways.

Queen of spade can be chosen from pack in only `1` way.

So, probability of choosing queen of spade = `1 /(52)`

Correct Answer is `=>` (A) `1 /(52)`

Q 2376191076

What is the probability that a leap year selected at
random contains 53 Mondays?
NDA Paper 1 2012

(A)

`1/7`

(B)

`2/7`

(C)

`7/366`

(D)

`26/183`

Solution:

We know that, a leap year contains 366 days in which
52 weeks and rest 2 days. Now, here 2 days are arranging like
that
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday),
(Wednesday, Thursday,) (Thursday, Friday), (Friday, Saturday),
(Saturday, Sunday).
:. Total sample events `n(S) = 7`
and number of favourable events, `n(E) = 2`

`therefore ` Required probability ` = (n(E))/(n(S)) = 2/7`

Correct Answer is `=>` (B) `2/7`

Q 2167480385

What is the probability of `5` Sunday in the month of
December ?
NDA Paper 1 2016

(A)

`1/7`

(B)

`2/7`

(C)

`3/7`

(D)

None of these

Solution:

Number of days in December `= 31`

`:.` Number of complete weeks `= 4 (i.e. 7 xx 4 = 28` days)

Remaining 3 days can be

`(M, T, W),(T, W, Th), (W, Th, F), (Th, F, Sa), (F, Sa, S),`

`(Sa, S, M), (S, M, T)` .

Out of these `7, 3` are favourable outcomes.

So, probability of having 5 Sunday in the month of

december `3/7`.

Correct Answer is `=>` (C) `3/7`

Q 1753556444

Number `X` is randomly selected
from the set of odd numbers and `Y` is randomly selected from
the set of even numbers of the set `{1, 2, 3, 4, 5, 6, 7}`. Let,
`Z = (X+ Y)`.

What is `P( Z = 5)` equal to?
NDA Paper 1 2014

(A)

`1/2`

(B)

`1/3`

(C)

`1/4`

(D)

`1/6`

Solution:

Given that,

`X =` Set of odd numbers from the set `A`.

`Y =` Set of even numbers from the set `A`.

Let set `A = {1, 2, 3, 4, 5, 6, 7}` and `Z = X + Y`

Now, `Z = 5` is only possible

when `X = 1, 3` and `Y = 4, 2`

Sample space `= {(1 , 2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2),

(5,4), (5,6), (7,2), (7,4), (7,6)}`

`:.` Total number of sample space, `n (s) = 12`

and favourable space `= { (1 ,4), (3,2)}`

`:.` Total number of favourable cases, `n (E) = 2`

So, ` p (z = 5) = (n(E_1))/(n(S)) = 2/(12) = 1/6`

Correct Answer is `=>` (D) `1/6`

Q 2222101931

If `x in [0,5]`, then what is the probability that
`x^ 2 -3x + 2 >= 0 ?`
NDA Paper 1 2015

(A)

`4/5`

(B)

`1/5`

(C)

`2/5`

(D)

`3/5`

Solution:

Total number of solutions, `n(S) = 5`

and favourable events are `3, 4, 5` .

`:. n(E)=3`

Hence `P(E) = (n(E))/(n(S)) = 3/5`

Correct Answer is `=>` (D) `3/5`

Q 2470080816

A cricket team plays `x` number of matches in
winter and wins m matches. Further, it plays
`y` number of matches in summer and wins
`n` matches. Its winning probability in both
the seasons is
UPSEE 2015

(A)

`m/x * n/y`

(B)

`x/m - y/n`

(C)

`(m+n)/(x+y)`

(D)

`(x+y)/(m+n)`

Solution:

Total number of matches in both seasons `= x + y`

Number of winning matches `= m + n`

`:.` Required probability `=(m+n)/(x+y)`

Correct Answer is `=>` (C) `(m+n)/(x+y)`

Q 1750701614

A fair coin is tossed four times. What is the probability
that at most three tails occur?
NDA Paper 1 2014

(A)

`7//8`

(B)

`15//16`

(C)

`13//16`

(D)

`3//4`

Solution:

Let `S` be the sample space of the experiment

and `E`. be the event that at most three tails occur.

Clearly, `n(S) = 2^4 = 16`

and `n(E) = text()^(4)C_0 + text()^(4)C_1 + text()^(4)C_2 + text()^(4)C_3`

` = 1 + 4 + ( 4 xx 3)/(2 xx 1) + 4 = 1 + 4 + 6 + 4 =15`

`:. P(E) = (n(E))/(n(S)) = (15)/(16)`

Correct Answer is `=>` (B) `15//16`

Q 2282312237

What is the probability that the sum of any two different
single digit natural numbers is a prime number?
NDA Paper 1 2015

(A)

` 5/(27)`

(B)

` 7/(18)`

(C)

`1/3`

(D)

None of these

Solution:

We have, single digit natural numbers `= (1, 2, 3, ... , 9)`

`:. n(S) = text()^(9)C_2 = 36 `

The sum of two different numbers which are prime `2, 3,
5, 7, 11, 13, 17, 19.`

i.e. `n(E) = (1, 2), (1, 4), (1, 6), (2, 3), (2, 5), (2, 9), (3, 2),`

`(4, 1), (6, 1), (9, 2), (7, 4), (4, 7), (7, 6), (6, 7)`

`= 14`

Hence, `P(E) = (n(E))/(n(S)) = (14)/(36) = 7/(18)`

Correct Answer is `=>` (B) ` 7/(18)`

Q 2387301287

In a class of 125 students 70 passed in
Mathematics, 55 passed in Statistics and
30 passed in both subjects. What is the
probability that a student selected at random from
the class has passed in only one subject'?
NDA Paper 1 2011

(A)

`13/25`

(B)

`3/25`

(C)

`17/25`

(D)

`8/25`

Solution:

The number of students that exactly pass in
Mathematics` = 70 - 30 = 40`
and the number of students that exactly pass in Statistics
`=55- 30 = 25`

`therefore` required probability ` = (40+25)/(125) = 65/125 = 13/25`

Correct Answer is `=>` (A) `13/25`

Addition and multiplication theorem of probability

Q 2731156922

A students appears for tests `I , II` and `III` . The students is considered successful if he passes in tests `I , II` or `I , III` or all the three. The probabilities of the student passing in tests `I , II` and `III` are `m , n` and `1/2` respectvely If the probability of the students to be successful is `1/2` then which one of the following is correct ?
NDA Paper 1 2016

(A)

`m(1+n) = 1`

(B)

`n(1+m) = 1`

(C)

`m = 1`

(D)

`m n = 1`

Solution:

P( the students to be successful is) = `1/2`

if he passes in tests `I , II` or `I , III` or all the three

`mn + m. 1/2 + m.n. 1/2 = 1/2`

`m(1+n) = 1`

Correct Answer is `=>` (A) `m(1+n) = 1`

Q 2107180088

A coin is tossed three times. What is the probability of
getting head and tail alternately ?
NDA Paper 1 2016

(A)

`1/8`

(B)

`1/4`

(C)

`1/2`

(D)

`3/4`

Solution:

Given, a coin is tossed three times.

Now, required probability `= P (HTH) + P(THT)`

`= 1/2 xx 1/2 xx 1/2 + 1/2 xx 1/2 xx 1/2 quad [∵ P(H) = P(T) = 1/2]`

` = 1/8 + 1/8 = 2/8 = 1/4 `

Correct Answer is `=>` (B) `1/4`

Q 2316180970

An um contains one black ball
and one green ball. A second um contains one white and one
green ball. One ball is drawn at random from each urn.

What is the probability of getting atleast one green
ball?
NDA Paper 1 2012

(A)

`1/2`

(B)

`1/3`

(C)

`1/4`

(D)

`3/4`

Solution:

Probability of getting atleast one green ball
= P(one green ball from urn I and one white ball from urn II)+ P (one
green ball from urn I and one green ball from urn If)+ P (one black
ball from urn I and one green ball from urn II)

` = 1/2xx1/2+1/2xx1/2+1/2xx1/2 = 3/4`

Correct Answer is `=>` (D) `3/4`

Q 2826445371

Two balls are selected from a box containing `2` blue and `7` red balls. What is the probability that atleast one ball is blue?

(A)

`2/9`

(B)

`7/9`

(C)

`5/(12)`

(D)

`7/(12)`

Solution:

Required probability `= P` (one ball is blue) `+`

`P` (both balls are blue)

` = 2/9 xx 7/8 + 2/9 xx 1/8 = (14)/(72) + 2/(72) = (16)/(72) = 2/9`

Correct Answer is `=>` (A) `2/9`

Q 2876245176

A box contains `6` distinct dolls. From this box, three dolls are randomly selected one by one with replacement. What is the probability of selecting `3` distinct dolls?

(A)

`5//54`

(B)

`12//25`

(C)

`1//20`

(D)

`5//9`

Solution:

`∵` Favourable event for first doll `= 1`

`:.` Probability for first doll `= 1//6`

Favourable event for second doll `= 5`

`:.` Probability for second doll `= 5//6`

Favourable event for third doll ` = 4`

`:.` Probability for third doll `= 4//6`

`:.` Required probability `= 1/6 · 5/6 . 4/6 = 5/(54)`

Correct Answer is `=>` (A) `5//54`

Q 2347301283

A husband and wife appear in an interview for two
vacancies in the same post. The probability of
husband's selection is `1/5` and that of wife's

selection is `1/3` What is the probability that only

one of them will be selected?
NDA Paper 1 2011

(A)

`1/5`

(B)

`2/5`

(C)

`3/5`

(D)

`4/5`

Solution:

Probability of selection of husband, `P(H) = 1/5`

`therefore P(barH)= 1-1/5 = 4/5`

and probability of selection of wife `P(W) = 1/3`

`therefore P(barW) = 1-1/3 = 2/3`

`therefore` Probability that only one of them is selected

`= P(H)P(barW)+P(barH)P(W)`

` = (1/5)(2/5)+(4/5)(1/3)`

`= 2/15+4/15 = 6/15 = 2/5`

Correct Answer is `=>` (B) `2/5`

Dependent and Independent event

Dependent and Independent event :

Q 2826456371

If `A` and `B` are two independent events such that `P (A) = 1 // 2` and ` P (B) = 1/5` , then
`P(A cup B)` is equal to

(A)

`1/5`

(B)

`2/5`

(C)

`3/5`

(D)

`4/5`

Solution:

Since, A and B are independent

events, therefore

`P(A cap B) = P(A) - P(B) = 1/2 xx 1/5 = 1/(10)`

Now, `P(A cup B) = P(A) + P(B)`

`- P (A cap B) = 1/2 + 1/5 - 1/(10) = 3/5`

Correct Answer is `=>` (C) `3/5`

Q 2846734673

A speaks truth in `60%` cases and B speaks truth in `70%` cases. The probability that they will say the same thing while describing single event, is

(A)

`0.56`

(B)

`0.54`

(C)

`0.38`

(D)

`0.94`

Solution:

Given, `P (A) = 0.6, P (B) = 0.7`

Here, A and B are independent events.

`:. P (A cap B) = P(A) xx P (B)`

`= 0.6 xx 0.7 = 0.42`

`P ( bar A cap bar B) = P (bar A) xx P (bar B)`

`= 0.4 xx 0.3 = 0.12`

Probability that A and B will say same

thing = Probability that both speak

truth or false

`= P(A cap B)+ P(bar A cap bar B)`

`= 0.42 + 0.12 = 0.54`

Correct Answer is `=>` (B) `0.54`

Q 2866634575

A and B are two events such that `P(A) = 0.3` and `P(A cup B) = 0.8`. If A and B are independent, then `P(B)` is

(A)

`2//3`

(B)

`3//8`

(C)

`2//7`

(D)

`5//7`

Solution:

`P(A cup B) = 0.8`

`=> P(A) + P(B) - P(A cap B) = 0.8`

`=> P (A) + P (B) - P (A) · ( B) = 0.8`

[ ∵ A and B are independent events]

`=> 0.3 + P(B) { 1 - P(A)} = 0.8`

`=> P (B) ( 1 - 03 ) = 05`

`=> 0.7 P (B) = 0.5`

`:. P(B ) = (0.5)/(0.7) = 5/7`

Correct Answer is `=>` (D) `5//7`

Q 2682801737

Probabilities of solving problem independently by `A` and `B` are `1/2` and `1/3` respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.
CBSE-12th 2011

Solution:

The probability of solving the problem independently by A and B are given as `1/2` and `1/3`

respectively.

i.e. `P(A) = 1/2 , P(B) = 1/3`.

`:. P(A cap B) = P(A).P(B)`

[Since the events corresponding to A and B are independent]

` = 1/2 xx 1/3 = 1/6`

(i) Probability that the problem is solved

`= P (A cup B)`

`= P (A) + P (B) - P (A cap B)`

`= 1/2 + 1/3 - 1/6`

` = ( 3 + 2 - 1)/6`

` = 4/6`

` = 2/3`

Thus, the probability that the problem is solved is `2/3`.

(ii) Probability that exactly one of them solves the problem

`= P (A - B) + P (B - A)`

` = [P (A) - P (A cap B) + [ P (B) - P (A cap B)]]`

` = (1/2 - 1/6) + (1/3 - 1/6)`

` = (3 - 1 + 2 - 1)/6`

` = 3/6`

` = 1/2`

Thus, the probability that exactly one of them solves the problem is ` 1/2`

Q 1746856773

If `A` and `B` are two independent events with `P(A) = 3/5` and `P(B) = 4/9`, then `P(A' cap B')` equals to
NCERT Exemplar

(A)

`4/(15)`

(B)

`8/(45)`

(C)

`1/3`

(D)

`2/9`

Solution:

`P(A' cap B') = 1 - P(A cup B)`

= `1 - [P(A) + P(B) - P(A cap B)]`

= `1 - [3/5 + 4/9 - 3/5 xx 4/9]` [`therefore P(A cap B) = P(A) . P(B)`]

= `1 - [(27 + 20 - 12)/(45)] = 1 - (35)/(45) = (10)/(45) = 2/9`

Correct Answer is `=>` (D) `2/9`

Mutually exclusive and exhaustive event

`P(A cup B) = 1 - bar (P(A cup B))`

For Mutually exclusive events

`P(A cup B) =P(A) + P(B) ` [`P (A cap B ) = 0`]

For Mutually exhaustive event

`P(A cup B) = 1`

For Mutually exclusive and exhaustive event

`P(A cup B) =P(A) + P(B) = 1`

Q 2733191942

Consider the following statements :

1. Two events are mutually exclusive if the occurrence of one event prevents the occurrence of the other.

2. The probability of the union of two mutually exclusive events is the sum of their individual probabilities.

Which of the above statements is/are correct?
NDA Paper 1 2017

(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

For mutually exclusive

`P(A cap B) =0`

Probability of occuring of `A` and `B=0`

`P(A cup B) =P(A) + P(B) +P(A cap B)`

`P(A cup B) = P(A) + P(B)`

Hence `2` is correct

`=> ` Hence A & B will be dependent event so they will prevent the occurance of each other.

Correct Answer is `=>` (C) Both 1 and 2

Q 2724101051

Let A and B are two mutually exclusive events with `p(A) = 1/3` and `p(B) = 1/4` .What is the value of `P(bar A cap bar B)`?
NDA Paper 1 2017

(A)

`1/6`

(B)

`1/4`

(C)

`1/3`

(D)

`5/12`

Solution:

`P(A) = 1/3`

`P(B) = 1/4`

`P(bar A cup bar B) =P(bar A)+ P(bar B) - P(bar A cap bar B)`

`1= 2/3 + 3/4 - P(bar A cap bar B)`

`P(bar A cap bar B) = 17/12 -1 = 5/12`

Correct Answer is `=>` (D) `5/12`

Q 2316878770

Consider the following statements
I. If A and B are exhaustive events, then their
union is the sample space.
II. If A and B are exhaustive events, then their
intersection must be an empty event.
Which of the above statements is/are correct?
NDA Paper 1 2013

(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

I. Two or mom events associated to a random
experiments are exhaustive, if their union is the sample space

`A_1 uu A_2 uu .........uuA_n = S`

`=> P(A_1 uu A_2 uu ........... uu A_n ) = 1`

II. Two or more events associated to a random experiment are
said to be mutually exclusive events, it the occurrence of one
of them prevents or denies the occurrence of all others

`E_1 nn E_2 nn ................ nn E_n = phi`

`P(E_1 nn E_2 nn ................... nn E_n) = 0`

Correct Answer is `=>` (A) Only 1

Q 2753891744

A question is given to three students A, B and C whose chances of solving it are
`1/2, 1/3` and `1/4`. respectively. What is the probability that the question will be solved?
NDA Paper 1 2017

(A)

`1/24`

(B)

`1/4`

(C)

`3/4`

(D)

`23/24`

Solution:

`P= 1-` (Question will be unsolved)

`=1- (1-1/2) (1-1/3) (1-1/4)`

`=1- 1/2 xx 2/3 xx 3/4`

`=3/4`

Correct Answer is `=>` (C) `3/4`

Q 2771156926

A salesman has a 70% change to sell a product to any customer .The behaviour of successive customers is independent . If two customers A and B enter what is the probablity that the salesman will sell the product to customer A or B ?
NDA Paper 1 2016

(A)

`0.98`

(B)

`0.91`

(C)

`0.70`

(D)

`0.49`

Solution:

P(sell) = .7

`P`(the salesman will sell the product to customer A or B `) = 1 - P(`the salesman will not sell the product to any customer)

`P = 1-(1-.7)(1-.7)`
`=0.91`

Correct Answer is `=>` (B) `0.91`

Q 2701456328

A machine is known to be `75%` effective to cure a patient. If the medicine is given to 5 patients, what is the probability that at least one patient is curved by this medicine ?
NDA Paper 1 2016

(A)

`1/1024`

(B)

`243/1024`

(C)

`1023/1024`

(D)

`781/1024`

Solution:

P(cure) `= 3/4`

P( atleast one is cured) `= 1-` None cured

`=1- (1-3/4) xx (1-3/4) xx (1-3/4) xx (1-3/4) xx (1-3/4)`

`= 1-(1/4)^5= 1023/1024`

Correct Answer is `=>` (C) `1023/1024`

Q 1689045817

If `A` and `B` are two events such that `P(A cup B) = 3/4`
`P(A cap B) =1/4` and `P(barA) = 2/3`, then what is `P(B)`
equal to?
NDA Paper 1 2015

(A)

`1/3`

(B)

`2/3`

(C)

`1/8`

(D)

`2/9`

Solution:

We have, `P (A cup B)= 3/4 P(A cap B)= 1/4 ` and `P(barA) =2/3`

To find `P(B)`

We know that,

`P (A cup B) = P(A) + P(B)- P(A cap B)`

`=> 3/4 = (1- P(barA)) + P(B)- 1/4`

`=> 3/4 + 1/4 = 1 - P(barA) + P(B) => 1 = 1 - P(barA) + P(B)`

`=> P(B) = P(barA) = 2/3`

Correct Answer is `=>` (B) `2/3`

Q 2724101051

Let A and B are two mutually exclusive events with `p(A) = 1/3` and `p(B) = 1/4` .What is the value of `P(bar A cap bar B)`?
NDA Paper 1 2017

(A)

`1/6`

(B)

`1/4`

(C)

`1/3`

(D)

`5/12`

Solution:

`P(A) = 1/3`

`P(B) = 1/4`

`P(bar A cup bar B) =P(bar A)+ P(bar B) - P(bar A cap bar B)`

`1= 2/3 + 3/4 - P(bar A cap bar B)`

`P(bar A cap bar B) = 17/12 -1 = 5/12`

Correct Answer is `=>` (D) `5/12`

Q 2733191942

(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Correct Answer is `=>` (C) Both 1 and 2

Q 2282101937

A problem in Statistics is given to three students `A, B` and
`C` whose chances of solving it independently are `1/2, 1/3` and
`1/4` respectively. The probability that the problem will be
solved, is
NDA Paper 1 2015

(A)

`1/(12)`

(B)

`(11)/(12)`

(C)

`1/2`

(D)

`3/4`

Solution:

`P` (Problem will be solved)

`= 1 - P` (problem will not solved by A, Band C)

`= 1 - { ( 1- 1/2 ) ( 1 - 1/3) (1 - 1/4) }`

` = 1 - 1/2 xx 2/4 xx 3/4 = 1 - 1/4 = 3/4`

Correct Answer is `=>` (D) `3/4`

Q 1700101018

Consider events `A, B, C, D` and `E` of the sample space
`5 = {n : n` is an integer such that `10 <= n <= 20)` (given)
A is the set of all even numbers.
B is the set of all prime! numbers.
`C = 15`
D is the set of all integers `<= 16`.
E is the set of all double digit numbers expressible as a
power of `2`.
`5 = {n: n` is an integer such that `10 <= n <= 20}`
`= {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}`
`A = {10, 12, 14, 16, 18, 20}`
`B = {11, 13, 17, 19}`
`C = {15}`
`D = {10, 11, 12, 13, 14, 15, 16}`
`E = {16}`

`C` and `E` are
NDA Paper 1 2014

(A)

mutually exclusive events but not elementary events

(B)

exhaustive events but not mutually exclusive events

(C)

mutually exclusive and exhaustive events

(D)

elementary and mutually exclusive events

Solution:

`C` and `E` are mutually exclusive and

elementary events.

Correct Answer is `=>` (D) elementary and mutually exclusive events

Q 1710701610

Two students `X` and `Y` appeared in an examination. The
probability that `X` will qualify the examination is `0.05`
and `Y` will qualify the examination is `0.10`. The
probability that both will qualify the examination is
`0.02`. What is the probability that only one of them will
qualify the examination?
NDA Paper 1 2014

(A)

`0.15`

(B)

`0.14`

(C)

`0.12`

(D)

`0.11`

Solution:

Let `A` and `B` be the events that `X` and `Y` qualify

the examination, respectively.

We have, `P(A) = 0.05, P(B) = 0.10` and `P(A cap B)= 0.02`

To find `P` (only one of `A` and `B` will qualify examination)

Clearly, `P` (only one of `A` and `B` will qualify the examination)

`= P(A cap bar B)+ P(B cap bar A)`

`= P(A) - P(A cap B)+ P(B)- P(A cap B)`

`= P(A) + P(B) - 2P (A cap B)`

`= 0.05 + 0.1- 2(0.02)`

`= 0.15 - 0.04 = 0.11`

Correct Answer is `=>` (D) `0.11`

Q 2377623586

NDA Paper 1 2009

Assertion : (A) The probability of drawing either an ace or a king from a deck of card in a single draw is `2/13`

Reason : (R) For two events `E_1` and `E_2` which are not mutually exclusive, the probability is given by `P(E_1+E_2) = P(E_1)+P(E_2)-P(E_1 nn E_2)`

(A)

Both A and R individually true and R is the correct explanation of A

(B)

Both A and R are individually true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Solution:

Required probability = `4/52+4/52 = 1/13+1/13 = 2/13`

and when two events are not mutually exclusive, then

`P(E_1+E_2) = P(E_1)+P(E_2)-P(E_1 nn E_2)`

Both A and R are true but R is not the correct explanation of A.

Correct Answer is `=>` (B)

Conditional Probability

Q 2783891747

For two dependent events A and B, it is given that `P(A) = 0.2` and `P(B) = 0·5`. If
`A subseteq B`, then the values of conditional probabilities `P(A|B)` and `P(B|A)` are
respectively
NDA Paper 1 2017

(A)

`2/5 , 3/5`

(B)

`2/5, 1`

(C)

`1, 2/5`

(D)

Information is insufficient

Solution:

`p(A) =0.2, P(B) =0.5`

If `A subseteq B , P(A cap B )= P(A)`

`p(A/B)= (P(A cap B))/(P(B)) =(P(A))/(P(B))= 2/5`

`P(B/A) =(P(A cap B))/(P(A)) =(P(A))/(P(A))=1`

Correct Answer is `=>` (B) `2/5, 1`

Q 2741456323

For two events A and B it is given that `P(A) = 3/5 , P(B) = 3/10` and `P(A/B) = 2/3`. If `barA` and `barB` are the complementary events of A and B then what is `P ( barA/ barB)` equal to ?
NDA Paper 1 2016

(A)

`3/7`

(B)

`3/4`

(C)

`1/3`

(D)

`4/7`

Solution:

`P(A/B) =(P (A cap B))/(P(B)) = 2/3`

`P(A cap B)= 2/3 xx 3/10 = 1/5`

`P(A cap B) = 1/5 = 2/10`

`P((bar A)/(bar B)) = (P(bar A cap bar B))/(P bar B) =(bar (P (A cup B)))/(P ( bar B)) =(1- P(A cup B))/(7/10)`

`=(1 - (3/5 + 3/10 - 2/10))/(7/10)`

`= (1 - 7/10)/(7 /10) =(3/10)/(7/10) = 3/7`

Correct Answer is `=>` (A) `3/7`

Q 2262101035

Two events A and B are such that `P` (not `B`) `= 0.8`,
`P(A cup B)= 0.5` and `P(A // B)= 0.4`. Then, `P(A)` is equal to
NDA Paper 1 2015

(A)

`0.28`

(B)

`0.32`

(C)

`0.38`

(D)

None of these

Solution:

Given, `P(bar B) = 0.8 :.P(B) = 1-P(bar B) = 1-0.8 = 0.2`

`P(A cup B) = 0.5, P(A // B)= 0.4`

`:. P(A // B) = (P(A cap B))/(P(B))`

` => 0.4 = (P(A cap B))/ ( 0.2)`

` => 0.08 = P (A cap B)`

Hence, `P(A cup B) = P(A) + P(B) - P(A cap B)`

`=> 0.5 = P(A) + 0.2 - 0.08`

`=> 0.5= P(A) + 0.12 => P(A) = 0.38`

Correct Answer is `=>` (C) `0.38`

Q 2137480382

For two mutually exclusive events `A` and `B, P(A) = 0.2`
and `P(barA cap B) = 0.3`. What is `P(A | (A cup B))` equal to?
NDA Paper 1 2016

(A)

`1/2`

(B)

`2/5`

(C)

`2/7`

(D)

`2/3`

Solution:

As, `bar A cap B = B - A cap B`

So, in given case, `P(bar A cap B) = P(B) = 0.3`

[` ∵ A` and `B` are mutually exclusive, so `A cap B = phi`

`=> P (A cap B) = 0 ]`

and `P(A | A cup B) =( P[A cap (A cup B) ])/(P(A cup B))`

`= (P(A))/(P(A) + P(B)) = (0.2)/(0.2 + 0.3) = 2/5`

Correct Answer is `=>` (B) `2/5`

Q 2317401380

If A and B are events such that `P(A uu B) = 0.5`,
`P(barB) = 0.8` and `P(A / B)= 0.4`, what is `P(A nn B)`
equal to' ?
NDA Paper 1 2011

(A)

`0.08`

(B)

`0.02`

(C)

`0.8`

(D)

`0.2`

Solution:

`because P(A uu B) = 0.5`, `P(barB) = 0.8` `P(A / B)= 0.4`

Here `P(B) = 1-P(barB) = 1-08 = 0.2`

Now `P(A/B) = ((A nn B))/(P(B))`

`=> P(B) xx P(A/B) = P( A nn B)` [ `because P(B) = 1-P(barB)`]

`=> P(A nn B) = 0.4xx(1-0.8)`

` = 0.4xx0.2 = 0.08`

Correct Answer is `=>` (A) `0.08`

Q 2573478346

If `A` and `B` are independent events such that
`P(A) > 0` and `P(B) > 0`, then
UPSEE 2011

(This question may have multiple correct answers)

(A) `P(A cup B) = P(A) * P(B)`

(B) `P(A cap B) = P(A) * P(B)`

(D) `P(B|A) = P(B)`

Solution:

Since, `A` and `B` are independent events.

`:. P(A cap B) = P(A) P (B)`

`P (A | B) = (P (A cap B))/(P(B)) = P(A)`

and `P(B | A) = (P (B cap A ) )/(P(A)) = P(B)`

Correct Answer is `=>` (B)

Q 2327623581

If `P(A)= 1/3, P(B) = 1/4, P(A/ B) = 1/6`, then
what is `P(B / A)` equal to?
NDA Paper 1 2009

(A)

`1/4`

(B)

`1/8`

(C)

`3/4`

(D)

`1/2`

Solution:

Given `P(A)= 1/3, P(B) = 1/4, P(A/ B) = 1/6`

But `P(A/B) = (P(A nn B))/(P(B))` (`because` by conditional probability)

`=> 1/6 = (P(A nn B))/(1/4)`

`=> P(A nn B) = 1/24`

`therefore P(A/B) = (P(A nn B))/(P(A))` (`because` by conditional probability)

`= (1/24)/(1/3) = 1/8`

Correct Answer is `=>` (B) `1/8`

Q 1619845710

If `A subseteq B`, then which one of the following is not
correct?
NDA Paper 1 2015

(A)

`P(A cap bar(B)) = 0`

(B)

`P(A | B)= (P(A))/(P(B))`

(C)

`P(B | A)= (P(B))/(P(A))`

(D)

`P(A | (A cap bar(B)) = (P(A))/(P(B))`

Solution:

As, `A subseteq B ) `, then `A cup B = B` and `A cap B =A`.

Clearly, `P(A cap bar( B))= P (phi) = 0`

Now, `P (A/B) = (P(A cap B))/(P(A)) = (P(A))/ (P(B))`

`=> P(( A )/ ( A cup B) ) = P(A/B) = (P(A))/ (P(A))`

but `P (B/A) = (P(B cap A))/( P(A)) = (P(A))/ (P(A)) = 1`

Hence, option `(c)` is not correct.

Correct Answer is `=>` (C) `P(B | A)= (P(B))/(P(A))`

Q 2326191071

Two dice each numbered from
1 to 6 are thrown together. Let A and B be two events given by
A : even number on the first die
B : number on the second die is greater. than 4

What is the value of `P (A nn B)?`
NDA Paper 1 2012

(A)

`1/2`

(B)

`1/4`

(C)

`2/3`

(D)

`1/6`

Solution:

Let `S` be the sample space
`therefore` `n(S) = 36`
A : even number on the first die
`= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}`
B : number on the second die is greater than 4
`{(1' 5), (1. 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6) (5,5) , (5 ,6) ,(6 ,6)}`

`n(A) = 18 , n(B) = 12`

`therefore P(A) = (n(A))/(n(S)) = 18/36 = 1/2`

`P(B) = (n(B))/(n(S)) = 12/36 = 1/3`

Also `A nn B = { (2, 5), (2, 6), (4, 5), (4, 6), (6, 5), (6, 6)}`

`P(A nn B) = 6`

`=> n(A nn B) = (n(A nn B))/(n(S)) = 6/36 = 1/6`

Correct Answer is `=>` (D) `1/6`

Problems Based on Dice And Ball

Q 2147180083

Three dice are thrown simultaneously. What is the
probability that the sum on the three faces is at least `5`?
NDA Paper 1 2016

(A)

` (17)/(18)`

(B)

` (53)/(54)`

(C)

` (103)/(108)`

(D)

` (215)/(216)`

Solution:

Total outcomes `= 6 xx 6 xx 6 = 216`

Now, required probability `= 1 - P ` (sum = `3` and `4`)

`= 1 - [ 1/(216) + 3/(216)]`

[`∵` sum `3 = {1, 1, 1}` and

sum ` 4 = {(1, 1, 2),(1, 2, 1), (2, 1, 1)}]`

`= 1 - 4/(216) = (216 - 4)/(216) = (212)/(216) = (53)/(54)`

Correct Answer is `=>` (B) ` (53)/(54)`

Q 2272101936

A bag contains `4` white and `2` black balls and another bag
contains `3` white and `5` black balls. If one ball is drawn
from each bag, then the probability that one ball is white
and one ball is black, is
NDA Paper 1 2015

(A)

`5/(24)`

(B)

`(13)/(24)`

(C)

`1/4`

(D)

`2/3`

Solution:

`P`(one ball is white and one ball is black) `=P`(black ball

from bag 1 and white ball from bag 2 or white ball from bag `1` and black ball from bag `2`)

` = 2/6 xx 3/8 + 4/6 xx 5/8 = (26)/(48) = (13)/(24)`

Correct Answer is `=>` (B) `(13)/(24)`

Q 1752480334

A box contains `3` white and `2` black balls. Two balls are
drawn at random one after the other. If the balls are not
replaced, what is the probability that both the balls are
black?
NDA Paper 1 2014

(A)

`2/5`

(B)

`1/5`

(C)

`1/10`

(D)

None of these

Solution:

Total sample space, `n(S) = text()^(5)C_2` because we

select two balls out of `5` balls.

Now, favourable events, `n(f) =` Two selected balls are

black.

` = text()^(3)C_0 xx text()^(2)C_2`

`:. ` Required probabilty `= (n(E))/(N(S)) = ( text()^(3)C_0 xx text()^(2)C_2)/( text()^(5)C_2)`

` = (1 xx 1)/((5 xx 4)/2) = 1/(10)`

Correct Answer is `=>` (C) `1/10`

Q 1720401311

There are `4` white and `3` black balls in a box. In another
box, there are `3` white and `4` black balls. An unbiased die
is rolled. If it shows a number less than or equal to `3`,
then a ball is drawn from the second box, otherwise from
the first box. If the ball drawn is black, then the
probability that the ball was drawn from the first box, is
NDA Paper 1 2014

(A)

`1//2`

(B)

`6//7`

(C)

`4//7`

(D)

`3//7`

Solution:

Box I `-> 4 W, 3 B`

Box II `-> 3 W, 4 B`

Probability for choosing first box `= 3/6 = 1/2`

Probability for choosing second box `= 1/2`

`:. ` Required probability `= (1/2 xx 3/7)/( 1/2 xx 3/7 + 1/2 xx 4/7)`

` = (3/(14))/( 3/(14) + 4/(14)) = (3//14)/(7//14) = 3/7`

Correct Answer is `=>` (D) `3//7`

Q 2701056828

Three candidates solve a question odds in favour of the correct answer are `5 : 2 , 4 : 3` and `3 : 4` respectively for the three candidates . What is the probability that at least two of them solve the question correctly ?
NDA Paper 1 2016

(A)

`209/343`

(B)

`134/343`

(C)

`149/343`

(D)

`60/343`

Solution:

odds in favour of the correct answer are `5 : 2 , 4 : 3` and `3 : 4`
probability in for correct answer are 5/7, 4/7, 3/7

P( that at least two of them solve the question correctly) = `Ist and II nd` do correct and `IIIrd` do incorrect + `Ist and III nd` do correct and `IIrd` do incorrect + `IIst and III nd` do correct and `IIrd` do incorrect

`P = 5/7 . 4/7 . (1-3/7) + 5/7 . 3/7 . (1-4/7) + 4/7 . 3/7 (1-5/7)`

`P = 209/343`

Correct Answer is `=>` (A) `209/343`

Q 2387501487

Two balls are selected from a box containing 2
blue and 7 red balls. What is the probability that
atleast one ball is blue'?
NDA Paper 1 2010

(A)

`2/9`

(B)

`7/9`

(C)

`5/12`

(D)

`7/12`

Solution:

Required probability = P (one ball is blue)

(both the balls are blue)

` = 2/9xx7/8+2/9xx1/8 = 14/72+2/72 = 16/72 = 2/9`

Correct Answer is `=>` (A) `2/9`

Probability using permutation and combination

Q 2876345276

Three digital numbers are formed using the digits `0, 2, 4, 6, 8.` A number is chosen at random out of these numbers. What is the probability that the number has the same digits?

(A)

`1//16`

(B)

`1//25`

(C)

`16//25`

(D)

`1//645`

Solution:

Favourable numbers `= {222, 444, 666, 888}`

Total digit numbers `= 4 xx 5 xx 5`

`:.` Required probability `= 4/(4 xx 25) = 1/(25)`

Correct Answer is `=>` (B) `1//25`

Q 2816145070

A cricket team has `15` members, of whom only `5` can bowl. If the names of the `15` members are put into a bat and `11` drawn at random, then the chance of obtaining an eleven containing atleast `3` bowlers is

(A)

`7//13`

(B)

`11//15`

(C)

`12//13`

(D)

`15//14`

Solution:

Required probability

` = (text()^5C_3 xx text()^(10)C_8)/(text()^(15)C_(11)) + (text()^5C_4 xx text()^(10)C_7)/(text()^(15)C_(11)) + (text()^5C_5 xx text()^(10)C_6)/(text()^(15)C_(11))`

`= 1/(text()^(15)C_(11)) ( 10 xx 45 + 5 xx 120 + 1 xx 210)`

` = ( 1260 xx 1 xx 2 xx 3 xx 4)/(15 xx 14 xx 13 xx 12) = (12)/(13)`

Correct Answer is `=>` (C) `12//13`

Q 2232101032

Three digits are chosen at random from `1, 2, 3, 4, 5, 6, 7, 8`
and `9` without repeating any digit. What is the probability that the product is odd ?
NDA Paper 1 2015

(A)

`2/3`

(B)

`7/(48)`

(C)

`5/(42)`

(D)

` 5/(108)`

Solution:

Here, `n(E) =text()^(9)C_3`

Let favourable event =`E`

`:. n(E) = text()^(5)C_3`

Now,` P(E) = (n(E))/(n(S)) = (text()^(5)C_3)/(text()^(9)C_3) = 5/(42)`

Correct Answer is `=>` (C) `5/(42)`

Q 2327501481

What is the probability of having 53 Sunday or
53 Monday in a leap year'?
NDA Paper 1 2010

(A)

`2/7`

(B)

`3/7`

(C)

`4/7`

(D)

`5/7`

Solution:

A leap year has 366 days, in which 2 days may be any
one of the following pairs :
(Sunday, Mcnday), (Monday, Tuesday), (Tuesday, Wednesday),
(Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday),
(Saturday, Sunday).
`therefore` Required probability = `2/7+2/7-1/7 = 3/7`

Correct Answer is `=>` (B) `3/7`

Q 2317301289

A box contains 6 distinct dolls. From the box,
3 dolls are randomly selected one by one with
replacement. What is the probability of selecting
3 distinct dolls'?
NDA Paper 1 2011

(A)

`5/54`

(B)

`12/25`

(C)

`1/20`

(D)

`5/9`

Solution:

Required probability = `(text()^6C_1xx text()^5C_1 xx text()^4C_1)/(text()^6C_1 xx text()^6C_1 xx text()^6C_1) = (6 *5*4)/(6*6*6) = 5/6`

Correct Answer is `=>` (D) `5/9`

Q 2733891742

A committee of two persons is constituted from two men and two women. What is the probability that the committee will have only women?
NDA Paper 1 2017

(A)

`1/6`

(B)

`1/3`

(C)

`1/2`

(D)

`2/3`

Solution:

`P= (2 M, 2W)`

We choose two women from community by `text()^2 C_2`

Required Probability `P= (text()^2 C_2)/(text()^4 C_2) = 1/6`

Correct Answer is `=>` (A) `1/6`

Q 2773091846

A card is drawn from a well-shuffled ordinary deck of 52 cards. What is the probability that it is an ace?
NDA Paper 1 2017

(A)

`1/13`

(B)

`2/13`

(C)

`3/13`

(D)

`1/52`

Solution:

We choose one Ace from four aces

`P=(text()^4 C_1)/(text()^52 C_1) = 1/13`

Correct Answer is `=>` (A) `1/13`

Q 2167601585

A fair coin is tossed `100` times. What is the probability of
getting tails an odd number of times?
NDA Paper 1 2016

(A)

`1/2`

(B)

`3/8`

(C)

`1/4`

(D)

`1/8`

Solution:

Clearly, total number of out comes `= 2^(100)` and number of

favourable outcomes

`= text()^( 100)C_(1) + text ()^(100)C_(3) + ... + text ()^(100)C_(99) = 2^(100-1)`

`[∵ text()^(n)C_(1) + text()^(n)C_(3) + text()^(n)C_(5) + ... = 2^(n-1)]`

`= 2^(99)`

Hence, required probability `= 2^(99)/ 2^(100) = 1/2 `

Correct Answer is `=>` (A) `1/2`

Q 2317401389

There are 4 letters and 4 directed envelopes. These
4 letters are randomly inserted into the
4 envelopes. What is ti{e probability that the
letters are inserted into the corresponding
envelopes'?
NDA Paper 1 2011

(A)

`11/12`

(B)

`23/24`

(C)

`1/24`

(D)

None of these

Solution:

Required probability `= 1/(4!) = 1/(4xx3xx2)`

`= 1/24`

Correct Answer is `=>` (C) `1/24`

Q 2377301286

In a random arrangement of the letters of the
word 'UNIVERSITY', what is the probability that
two I ' s do not come together?
NDA Paper 1 2011

(A)

`4/5`

(B)

`1/5`

(C)

`1/10`

(D)

`9/10`

Solution:

Total number of ways of the arrangement of the letters
of the word 'UNIVERSITY' `= (10!)/(2!)`

Number of ways that both l's come together `= 9!`
`therefore` Number of ways that both l's do not come together

`= (10!)/2-9!`

`therefore` Required probability ` = (((10!)/2-9!)/((10!)/2)) = 4/5`

Correct Answer is `=>` (A) `4/5`

Bay's theorem and Total probability theorem

Q 1609645518

Two similar boxes `B_i (i = 1, 2)` contain `( i + 1)` red and
`(5- i -1)` black balls. One box is chosen at random
and two balls are drawn randomly. What is the
probability that both the balls are of different
colours?
NDA Paper 1 2015

(A)

`1/2`

(B)

`3/(10)`

(C)

`2/5`

(D)

`3/5`

Solution:

We have, box `B _1,` containing `2` red and `3` black

balls and box `B_2` containing `3` red and `2` black balls.

Let `E_1` be the event that box `B_1` is chosen,

`E_2` be the event that box `B_2` is chosen and `A` be the

event that balls are of different colours.

Clearly, `P(E1) = P(E_2 ) = 1/2`

`P (A/E_1) = (text()^(2)C_1 xx text()^(3)C_1)/ (text()^(5)C_2)`

and `p (A/E_2 ) = (text()^(3)C_1 xx text()^(2)C_1)/ (text()^(5)C_2) = 6/(10) = 3/5`

Now, by the theorem of total probability, we have

`P (A) = P(E_1) . P(A/E_1 ) + P(E_2 ) . P(A/E_2)`

`= 1/2 . 3/5 + 1/2 . 3/5 = 1/2 ( (2 xx 3)/5) = 3/5`

Correct Answer is `=>` (D) `3/5`

Q 2282212137

An insurance company insured `2000` scooter drivers,
`4000` car drivers and `6000` truck drivers. The probabilities
of an accident involving a scooter driver, car driver and a
truck driver are `0.01, 0.03` and `0.15`, respectively. One of
the insured persons meets with an accident. The
probability that the person is a scooter driver, is
NDA Paper 1 2015

(A)

`1/(52)`

(B)

`3/(52)`

(C)

`(15)/(52)`

(D)

`(19)/(52)`

Solution:

Let `P(A) = P`(scooter) `= (2000)/(12000) = 1/6`

`P(B) = P`(car) `= (4000)/(12000) = 1/3`

and `P(C) =P`(truck) `= (6000)/(12000) = 1/2`

Let `E =` Event that person meets with accident.

Then, `P(E/A) = 1/(100) , P(E/B) = 3/(100) , P(E/C) = (15)/(100)`

Now, ` P(A/E) = (P(A). P(E/A))/( P(A). P(E/A) + P(B). P(E/B) + P(C). P(E/C))`

` = ( 1/6 xx 1/(100))/(1/6 xx 1/(100) + 1/3 xx 3/(100) + 1/2 xx (15)/(100) ) = (1/6)/(1/6 + 1 + (15)/2)`

` = (1/6)/((1+ 6 + 45)/6) = 1/(52)`

Correct Answer is `=>` (A) `1/(52)`

Q 2856134974

Of cigarette smoking population `70%` are men and `30%` are women, `10%` of these men and `20%` of these women smoke Wills. The probability that a person seen smoking a Wills to be men is

(A)

`1//5`

(B)

`7//13`

(C)

`5//13`

(D)

`7//10`

Solution:

Let `E_1` and `E_2` denote the cigarette

smoking population of men and women,

respectively.

Given, `P(E_1 ) = 0.7, P (E_2 ) = 03.`

` P( A/E_1) = 0.1 , P ( A/E_2) = 0.2`

Using Baye's theorm , ` P(E_1 // A)`

` = ( P (E_1) P (A // E_1))/(P (E_1) xx P(A // E_1) +P (E_2) xx P(A // E_2))`

` = (0.7 xx 0.1)/(0.7 xx 0.1 + 0.3 xx 0.2)`

` = (0.07)/(0.07 + 0.06) = (0.07)/(0.13) = 7/(13)`

Correct Answer is `=>` (B) `7//13`

Probability Distribution : Mean, Variance, Binomial distribution

Q 2703891748

A point is chosen at random inside a circle. What is the probability that the point is closer to the centre of the circle
than to its boundary?
NDA Paper 1 2017

(A)

`1/5`

(B)

`1/4`

(C)

`1/3`

(D)

`1/2`

Solution:

Let a circle of centre `(0,0)` and radius `r`

to be point closer to the centre of circle than its boundary point should lie in the area of a circle of radius `r/2`

Hence probability `P= (pi xx (r/2)^2)/(pi r^2)= 1/4`

Correct Answer is `=>` (B) `1/4`

Q 2147380283

A certain type of missile hits the target with probability
`p` . What is the least number of missiles should be
fired so that there is atleast an `80%` probability that the
target is hit?
NDA Paper 1 2016

(A)

`5`

(B)

`6`

(C)

`7`

(D)

None of these

Solution:

Let `n` missiles be fired and `r` of them hits the target.

`:. P(X = r) = text()^(n)C_(r) p^(r) q^(n -r)`

` = text()^(n)C _(r) (0.3)^(r) (0.7)^(n -r)`

Target is hit when atleast 1 missile strikes the target.

` :. P(X >= 1) = 1 - P( X = 0 )`

This must be greater than `80%.`

So `1- text()^(n)C_(0) (0.3)^(0) (0.7)^(n- 0) >= (80)/(100)`

` => 1 - (7/(10))^(n) >= (80)/(100)`

` => (7/(10))^(n) <= (20)/(100) => n >= 5`

Correct Answer is `=>` (A) `5`

Q 2187680587

A point is chosen at random inside a rectangle
measuring `6` inches by `5` inches. what is the probability
that the randomly selected point is atleast one inch from
the edge of the rectangle?
NDA Paper 1 2016

(A)

`2/3`

(B)

`1/3`

(C)

`1/4`

(D)

`2/5`

Solution:

Let `A_(2) =` Area in which a randomly selected point lies

`= (6- 2) xx (5-2)= 4 xx 3= 12` sq inch

`A_(1)` = Total area `= 6 xx 5 = 30` sq inch

`:.` Required probability `= A_(2)/A_(1) = (12)/(30) = 2/5`

Correct Answer is `=>` (D) `2/5`

Q 2202401338

Seven unbiased coins are tossed `128` times. In how many
throws would you find atleast three heads?
NDA Paper 1 2015

(A)

`99`

(B)

`102`

(C)

`103`

(D)

`104`

Solution:

Given, `p = q= 1/2 , n = 7, N = 128`

and ` r = 3, 4, 5, 6, 7`

`:. P(X >= 0) = 128 [ text()^7C_3 p^3q^4 + text()^7C_4 p^4q^3 + text()^7C_5 p^5q^2 + text()^7C_6 p^6q^1 + text()^7C_7 p^7q^0 ]`

` = 128 [ 35(1/2)^3(1/2)^4 + 35(1/2)^4 (1/2)^3 + 21(1/2)^5(1/2)^2 + 7(1/2)^6(1/2)^1 + (1/2)^7 ]`

` = 128 { (1/2)^7 [ 35 + 35 + 21 + 7 + 1 ] }`

` = [ 1/(128) xx 99 ] xx 128 = 99`

Correct Answer is `=>` (A) `99`

Q 1659745614

In an examination, the probability of a candidate
solving a question is `1/2` Out of given `5` questions in
the examination, what is the probability that the
candidate was able to solve at least `2` questions?
NDA Paper 1 2015

(A)

`1/(64)`

(B)

` 3/(16)`

(C)

`1/2`

(D)

`(13)/(16)`

Solution:

Let solving a question is success and not solving

a question is failure.

Let probability of success `= p = 1/2`

and probability of failure `= q = 1/2`

Let `X` be random variable that denotes the number of

success in `5` trials.

Clearly, `X - B` in `( 5, 1/2)`

`:. P (X = x) = text()^( 5)C_x(1/ 2) ^x (1/2)^(5-x)`

`=> x = 0, 1, 2, 3, 4, 5 = text()^( 5)C_x(1/ 2)^5`

Required probability `= P (X >= 2)`

` = 1 - [P (x = 0) + P (x = 1)]`

` = 1- (text()^( 5)C_0(1/ 2)^5 + text()^( 5)C_1(1/ 2)^5)`

` = 1- 6/(36) = (26)/(32) =(13)/(16)`

Correct Answer is `=>` (D) `(13)/(16)`

Q 2212212139

A coin is tossed `5` times. The probability that tail appears
an odd number of times, is
NDA Paper 1 2015

(A)

`1/2 `

(B)

` 1/3`

(C)

`2/5`

(D)

`1/5`

Solution:

We have,

`p =` Probability of tail appear `= 1/2`

`q = `Probability of tail not appear `= 1- p = 1/2`

Now, `P` (tail appear odd number of times)

`=P(X = 1) +P(X =3) +P(X =5)` `\ \ \ \ \ \ \ \ [∵ P(X =r)]`

`= text()^(n)C_r p^r q^(n -r)`

`=text()^( 5)C_1 (1/2)^2 (1/2)^4 + text()^( 5)C_3 (1/2)^3 (1/2)^2 + text()^( 5)C_5 (1/2)^5`

` 5/(32) + (10) /(32) + 1/(32) = (16)/(32) = 1/2`

Correct Answer is `=>` (A) `1/2 `

Q 2560667515

A fair coin is tossed at a fixed number of
times. If the probability of getting exactly `3`
heads equals the probability of getting
exactly `5` heads, then the probability of
getting exactly one head is
WBJEE 2015

(A)

`1/64`

(B)

`1/32`

(C)

`1/16`

(D)

`1/8`

Solution:

Let the coin be tossed `n` times.
Let getting head is consider to be success.

`:. p =1/2 , q =1-p = 1-1/2 = 1/2`

It is given that,

`P(X=3) = P(X = 5)`

`=> text()^(n)C_3 (1/2)^3 (1/2)^(n-3) = text()^(n )C_5 (1/2)^5 (1/2)^(n-5)`

`=> text()^(n)C_3 = text()^(n)C_5`

`=> n = 3+5 ` `[ :. text()^(n)C_x = text()^(n)C_y => x+y = n ]`

`=> n = 8`

Now, `P(X =1) = text()^(8)C_1 (1/2)^1 (1/2)^(8-1)`

`= text()^(8)C_1 xx (1/2)^8 = 1/32`

Correct Answer is `=>` (B) `1/32`

Q 2384023857

A student appears for tests `A, B` and `C`. He is successful if he passes either in
tests `A` and `B` or in tests `A` and `C`. The probabilities of the student passing tests

`A, B` and `C` are respectively `p, q` and `1/2`

If the probability that the student is
successful is `1/2`, then we can have
BITSAT Mock

(A)

`p = q = 1`

(B)

`p = q = 1/2`

(C)

` p = 0, q = 1`

(D)

infinitely many values of the pairs `(p, q)`

Solution:

`P` (the student passes)

`= P` (the student passes in tests `A` and `B` and fails in `C`) `+ P` (the student passes in tests `A` and `C` and fails in `B`) `+ P` (the student passes in all the three tests)

`= pq (1-1/2) + p 1/2 (1-q) +pq * 1/2`

`= p/2 (1+q)`

`:.` we must have `1/2 p (1+q) = 1/2`

`=> p (1+q) =1`

`:.` `p = 1, q = 0` is possible and also,

there are infinitely many values of

the pairs `(p, q)` connected by the

relation `p = 1/(1+q)`

Correct Answer is `=>` (D) infinitely many values of the pairs `(p, q)`

Q 2282501437

A coin is tossed five times. What is the probability that
heads are observed more than three times ?
NDA Paper 1 2015

(A)

` 3/(16)`

(B)

` 5/(16)`

(C)

`1/2`

(D)

` 3/(32)`

Solution:

Given, `p = P` (getting a head) = `1/2`

`q = P` (getting no head) `= 1/2`

`n = 4, r = 4` and `N = 5`

`:. P(X > 3) =5[ text()^(4)C_4 P^(4)q^(0)] = 5 [ 1xx (1/2)^4 ] = 5/(16)`

Correct Answer is `=>` (B) ` 5/(16)`

Q 1659645514

Two men hit at a target with probabilities `1/2` and `1/3`
respectively. What is the probability that exactly
one of them hits the target?
NDA Paper 1 2015

(A)

`1/2`

(B)

`1/3`

(C)

`1/6`

(D)

`2/3`

Solution:

Let the men's are Mr. `A` and Mr. `B`.

Let `A` be the event that Mr. `A` hit the target and `B` be

the event that Mr. `B` hit the target

`:. P(A) = 1/2 ` and `P(B) = 1/3`

Now,` P` (exactly one of them hits the target)

`= P (A cap barB ` or `bar A cap B) = P(A cap bar B)+ P (barA cap B)`

`= P(A) . P(bar B) + P(bar A) . P(B) = 1/2 . 2/3 + 1/2 . 1/3 = 3/6 = 1/2`

Correct Answer is `=>` (A) `1/2`

Q 1689845717

The mean and the variance in a binomial
distribution are found to be `2` and `1` respectively.
The probability .`P(X = 0)` is
NDA Paper 1 2015

(A)

`1/2`

(B)

`1/4`

(C)

`1/8`

(D)

`1/(16)`

Solution:

Let `X` be a random variable that follows binomial

distribution with parameter `n` and `p`.

Then, Mean `= E(x) = np = 2` ..........(1)

and variance `= var(x) = npq = 1` ...........(2)

On dividing Eq. (ii) by Eq. (i), we get `q = 1/2`

`=> p = 1/2 quad (∵ p + q = 1)`

Now, on substituting the value of `p` in Eq. (i), we get

`n = 4`

Thus, `X- B` in `(1/2,4)` and `P(X=x) = text()^( 4)C_x (1/2)^x (1/2)^(4-x)`,

`x = 0, 1, 2 ,3 ,4 = P(x = 0) = text()^(4)C_0 (1/2)^0 (1/2)^(4-0)`

`= text()^(4)C_0 (1/2)^4 = 1 xx 1/(16) = 1/(16)`

Correct Answer is `=>` (D) `1/(16)`

Q 2367301285

There is a point inside a circle. What is the probability that this point is close to the
circumference than to the centre?
NDA Paper 1 2011

(A)

`3/4`

(B)

`1/2`

(C)

`1/4`

(D)

`1/3`

Solution:

Let radius of given circle be `r` Now, make a concentric
circle with radius `r/2`

The given point is close to the circumference than to the centre, if
it lies in the shaded region

`therefore` Required probability = `(pi (r^2-(r/2)r^3))/(pir^r) = ((3/4)r^2)/r^2 = 3/4`

Correct Answer is `=>` (A) `3/4`

Q 2376080876

What is the most probable number of successes in
`10 ` trials with probability of success `2/3?`
NDA Paper 1 2012

(A)

`10`

(B)

`7`

(C)

`5`

(D)

`4`

Solution:

Given that, `n = 10, P` (success) `= 2/3`

Then, `q = 1-p = 1-2/3 = 1/3`

Now, by Binomial distnbution

`P(X = r) = text()^nC_r P^4 q^(n-r)` .............(i)

`P(X = 6) = text()^(10)C_6 xx (2/3)^6.(1/3)^4`

` = (10xx9xx8xx7xx)/(4xx3xx2xx1)(2^6)/(3^(10)) = (13440)/(3^(10))`

`P(X =7 ) = text()^(10)C_7 xx (2/3)^7xx(1/3)^3 = (10 * 9 * 8 )/(3*2*1) (2^7)/(3^(10)) = (15360)/(3^(10))`

`P(X = 8) = text()^(10)C_8 xx (2/3)^8(1/3)^2`

`=(10*9)/(2*1)(2^8)/(3^(10)) = (11520)/(3^(10))`

Clearly, the most probable number is 7.

Correct Answer is `=>` (B) `7`

Dependent And Independent Events

Q 2157180084

Two independent events `A` and `B` have `P(A) = 1/3` and
`P(B) = 3/4`. What is the probability that exactly one of the
two events `A` or `B` occurs?
NDA Paper 1 2016

(A)

`1//4`

(B)

`5//6`

(C)

`5//12`

(D)

`7//12`

Solution:

Given, `P(A) = 1/3` and `P(B) = 3/4`

Now, `P` (exactly one) `= P(A cup B) - P ( A cap B)`

`= P(A) + P(B) - P(A cap B) - P(A cap B)`

`[ ∵ P(A cup B) = P(A) + P(B) - P(A cap B) ]`

`= P(A) + P(B) - 2(P(A cap B))`

`= P(A) + P(B) - 2P(A) · P(B)`

[`∵` event, `A` and `B` are independent ]

`= 1/3 + 3/4 - 2 1/3 * 3/4 = 1/3 + 3/4 - 1/2 = (4 + 9 - 6)/(12) = 7/(12)`

Correct Answer is `=>` (D) `7//12`

Q 1730301212

For any two events `A` and `B`, which one of the following holds?
NDA Paper 1 2014

(A)

`P(A cap B) <= P(A) <= P(A cup B) <= P(A) + P(B)`

(B)

`P(A cup B) <= P(A) <= P(A cap B) <= P(A) + P(B)`

(C)

`P(A cap B) <= P(B) <= P(A cap B) <= P(A) + P(B)`

(D)

`P(A cap B) <= P(B) <= P(A) + P( B) < = P(A cup B)`

Solution:

Clearly, `A cap B subseteq A`

`=> P(A cap B) < = P(A)` ..........(1)

` A subseteq A cup B`

`=> P(A) <= P(A cup B)` ............(2)

We know that, `P(A cup B) = P(A) + P(B) - P(A cap B)`

`=> P(A cup B) <= P(A) + P(B)` ...........(3)

From Eqs. (1), (2) and (3),

`P(A cap B) <= P(A) <= P(A cup B) <= P(A) + P(B)`

Correct Answer is `=>` (A) `P(A cap B) <= P(A) <= P(A cup B) <= P(A) + P(B)`