Mathematics Must Do Problems Of Probability For NDA

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Must Do Problems Of Probability For NDA

Must Do Problems Of Probability

Must Do Problems

Q 2562791635

`4` boys and `2` girls occupy seats in a row at random. Then the probability that the two
girls occupy seats side by side is
WBJEE 2011

(A)

`1/2`

(B)

`1/4`

(C)

`1/3`

(D)

`1/6`

Solution:

Girls sit side by side, it means both girls are
seated together in `2!` ways.

Let `n(E) = 4` boys and `2` girls occupy seats
side by side

`= 5! 2!`

`n(S) = 6!`

`:.` Required probability `=(5! 2!)/(6!) = 2/6 =1/3`

Correct Answer is `=>` (C) `1/3`

Q 2806245178

An experiment consists of flipping a coin and then flipping it a second time if head occurs. If a tail occurs on the first flip, then a six-faced die is tossed once. Assuming that the outcomes are equally likely, what is the probability of getting one head and one tail?

(A)

`1//4`

(B)

`1/(36)`

(C)

`1//6`

(D)

`1//8`

Solution:

The favourable events to comes head at flipping a first coin

`= {H T,H H}`

The favourable events to comes tail at flipping a first coin

`= {T1, T2, T3, T4, T5, T6}`

The total favourable events

`= (HT, HH, T1, T2, T3, T4, T5, T6}`

`:.` Required probability `= 1/8`

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

Q 2573491346

If `A` and `B` are two events such that `P(A) = 3/4` and `P(B) = 5/8`, then
UPSEE 2010

(A)

`P(A cup B) ge 3/4`

(B)

`P(A' cap B) le 1/4`

(C)

`3/8 le P(A cap B) le 5/8`

(D)

All of the above

Solution:

`A ⊆ A cup B => P(A) le P(A cup B)`

`=> P (A cup B) ge 3/4`

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

`ge P(A) +P (B) -1`

`=3/4 +5/8 -1 =3/8`

Now, `A cap B ⊆ B => P(A cap B) le P(B) =5/8`

`:.3/8 le P(A cap B) le 5/8`.......(i)

Next `P(A cap B') = P(A) -P(A cap B)`

`=> 3/4 -5/8 le P(A cap B') le 3/4 - 3/8`

`=> 1/8 le P(A cap B') le 3/8`

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

`:. P(A cap B) = P(B) -P(A' cap B)` [using Eq. (i)]

`=> 3/8 le P(B) -P(A' cap B) le 5/8`

`=> 0 le P(A' cap B) le 1/4`

Correct Answer is `=>` (D) All of the above

Q 2816734679

The probability that in the toss of two dice, we obtain an even sum or a sum less than `5` is

(A)

`1//2`

(B)

`1//6`

(C)

`1//3`

(D)

`5//9`

Solution:

Let A = Event of getting an even Sum

`= { ( 1, 1), ( 1, 3), (3, 1), (2, 2),`

`( 1, 5), (5, 1), (2, 4), ( 4, 2), (3, 3),`

`(2, 6), ( 6, 2), (3, 5), (5, 3), ( 4, 4),`

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

and `B = `Event of getting sum less than `5`

`= { ( 1, 1) , ( 2, 1 ) , ( 1, 2),`

`( 1, 3), (3, 1), (2, 2)`

`=>A cap B = { ( 1, 1 ),(1, 3),(3, 1),(2, 2)}`

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

` = (18)/(36) + 6/(36) - 4/(36) = 5/9`

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

Q 2543780643

In a certain population 10% of the people are rich, 5% are famous and 3% are rich and famous. The probability that a person picked at random from the population is either famous or rich but not both, is equal to
UPSEE 2013

(A)

`0.07`

(B)

`0.08`

(C)

`0.09`

(D)

`0.12`

Solution:

`P(R) = P(text(Rich)) = 10% = 10/100 = 1/10`

`P(F) = P(text(Famous)) = 5% = 5/100 = 1/20`

`P ( R nn F) = P text(Rich and Famous) = 3% = 3/100`

`therefore` Probability of getting either rich or famous but not both

` = P ( R nn F')+P( R' nn F)`

` = P(R)-P(R nn F)+P(F)-P(R nn F)`

` = P(R)+P(F)-2P(R nn F)`

` = 1/10+1/20-6/100 = (10+5-6)/100`

` = 9/100 = 0.09`

Correct Answer is `=>` (C) `0.09`

Q 2806345278

What is the probability of having `53` Sundays or `53` Mondays in a leap year?

(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 anyone of the

following pairs.

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday),

(Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday),

(Saturday, Sunday).

`:.` Required probability `= 2/7 + 2/7 - 1/7 = 3/7`

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

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`

Q 2886145077

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 selection is `1 // 3`. What is the probability that only one of them will be selected ?

(A)

`1/5`

(B)

`2/5`

(C)

`3/5`

(D)

`4/5`

Solution:

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

`:. P (bar H) = 1 - 1/5 = 4/5`

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

`:. P ( bar W) = 1 - 1/3 = 2/3`

`:.` Probability that only one of then is selected

`= P(H) P(bar W) + P( bar H) P(W)`

`= (1/5) (2/3) + (4/5) (1/3) = 2/(15) + 4/(15) = 6/(15) = 2/5`

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

Q 2478034806

Two dice are thrown together. Then the
probability that the sum of numbers
appearing on them is a prime number, is
UPSEE 2009

(A)

` 5/(12)`

(B)

`7/(18)`

(C)

`(13)/(36)`

(D)

`(11)/(36)`

Solution:

The prime numbers between `2` to `12` are `2, 3, 5,`

`7, 11`.

Case I When sum is `2`, total cases are `(1, 1)` ie, `1`

Case II When sum is `3`, total cases are `(1, 2)`,

`(2, 1)` ie,` 2`

Case III When sum is `5`, total cases are `(1, 4)`,

`(2, 3), (4, 1), (3, 2)` ie, `4`

Case IV When sum is `7`, total cases are `(1, 6)`,

`(2, 5), (3, 4), (6, 1), (5, 2), (4, 3)` ie, `6`

Case V When sum is `11`, total cases are `(5, 6)`,

`(6, 5)` ie, `2`

`:.` Required probability `= (15)/(36) = 5/(12)`

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

Q 2826145071

The probability of India winning a test match against England is `1//2`. Assuming independence of the result of various matches, the chance that in a `5` match series, India's second win occur at `3`rd test is

(A)

`2/3`

(B)

`1/4`

(C)

`1/8`

(D)

`1/2`

Solution:

Required probability `= P` (India wins first and third test)

`+ P` (India wins second and third test)

`= 1/2 (1 - 1/2) (1/2) + (1 - 1/2) (1/2) (1/2) = 1/8 + 1/8 = 1/4`

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

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 1530145912

If `A` and `B` are events of a random experiment such that `Pbar((A cup B)) =4/5 , P( barA cup bar(B))= 7/10 ` = and `P(B) =2/5 ` , then `P(A)` equals
BITSAT 2009

(A)

` 9 /(10)`

(B)

` 8 /(10)`

(C)

` 7 /(10)`

(D)

` 3 /5`

Solution:

Given, `P (bar ( A) cup bar ( B))= P bar ((A cap B )) = 7 /10 `

Since, `P ( A cap B) + P bar ((A cap B )) = 1/3 `

`=> P( A cap B) = 1 - 7/10 = 3 /10 `

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

`=> 4/5 = P (A) + 2/5 - 3/10 `

`=> P (A) = 4/5 - 2/5 + 3/10`

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

Correct Answer is `=>` (C) ` 7 /(10)`

Q 2209545418

Two cards are drawn successively with
replacement from a pack of `52` cards. The
probability of drawing two aces is :
BITSAT Mock

(A)

`1/(12) xx 4/(51)`

(B)

`1/(52) xx 1/(51)`

(C)

`1/(13) xx 1/(13)`

(D)

`1/(13) xx 1/(17)`

Solution:

Let `S_i` denote the event of getting an ace

in the `i^(th)` draw.

`∴` Probability of getting aces in both the

draws

`= P (S_1 ∩ S_2) = P (S_1) P (S_2)`

[Multiplication theorem]

`= 4/(52) xx 4/(52)`

`= 1/(13) xx 1/(13)`

Correct Answer is `=>` (C) `1/(13) xx 1/(13)`

Q 2876534476

In shuffling a pack of cards `3` are accidently dropped, then the chance that missing card should he of different suits is

(A)

`169//425`

(B)

`261//425`

(C)

`104//425`

(D)

`425//169`

Solution:

Total ways `= text()^(52)C_3 = 22100`

There are `4` suits in a pack of cards, so

three suits can be selected in `text()4 C_3` ways

and one card each from different suits

can be selected in `text()^(13) C_1 xx text()^(13) C_1 xx text()^(13) C_1 `

ways.

So, favourabe ways

` text()4 C_3 xx text()^(13) C_1 xx text()^(13) C_1 xx text()^(13) C_1`

`= 8788`

`:.` Required probability ` = (8788)/(22100) = (169)/(425)`

Correct Answer is `=>` (A) `169//425`

Q 2886534477

If there are `4` addressed envelopes and `4` letters. Then, the chance that all the letters are not mailed through proper envelope is

(A)

`1//24`

(B)

`1`

(C)

`23//24`

(D)

`9//2`

Solution:

Sample space `= 4! = 4 xx 3 xx 2 xx 1 = 24`

If all letter mailed right addressed

envelopes, favourable outcomes

`= 1 xx 1 xx l xx 1 = 1`

`P(E ) = text( Favourable outcomes)/text( Total outcomes) = 1/( 24)`

The chance that all the letters not

mailed through proper envelope

`P(E') = 1 - P(E) = 1 - 1/(24) = (23)/(24)`

Correct Answer is `=>` (C) `23//24`

Q 2551034824

There are two coins, one unbiased with probaility `1/2` or getting heads and the other one is biased with probability `3/4` of getting heads. A coin is selected at random and tossed. It shows heads up. Then, the probability that the unbiased coin was selected is
WBJEE 2013

(A)

`2/3`

(B)

`3/5`

(C)

`1/2`

(D)

`2/5`

Solution:

Let `E ->` Event of head showing up

` E_1 ->` Event of biased coin chosen

` E_2 ->` Event of unbiased coin chosen

Now `p(E_2) = 1/2` and `p(E_1) = 1/2`

Also `p(E/E_2) = 1/2` and `p(E/E_1) = 3/4` (by conditional probability)

By Baye's theorem

`p(E_2/E) = ( P(E_2) * p(E/E_2))/(p(E_2) . p(E/E_2)+p(E_1).p(E/E_1))`

` = (1/2xx1/2)/(1/2xx1/2+1/2xx3/4) = 2/5`

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

Q 2468023805

Two dice are thrown `n` times in succession.
The probability of obtaining a double six at
least once is
UPSEE 2009

(A)

` (1/(36))^n`

(B)

` 1 - ((35)/(36))^n`

(C)

` (1/(12))^n`

(D)

None of these

Solution:

If p is the probability of success and q is

the probability of failure, then the probability of at

least one success in n trial is `1 - q^n`.

Here, p = Probability of getting double six in

two dice

`= 1/6^2 = 1/(36) ` and `q = (35)/(36)`

`:.` Required probability ` = 1` - (Probability of not

getting double six`)^n`

`= 1 - (( 35)/(36))^n`

Correct Answer is `=>` (B) ` 1 - ((35)/(36))^n`

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`

Q 2249678513

If from each of the three boxes containing `3` white and `1` black, `2` white
and `2` black, `1` white and `3` black balls, one ball is drawn at random, then the

probability that `2` white and `1` black ball will be drawn is
BITSAT Mock

(A)

`13/32`

(B)

`1/4`

(C)

`1/32`

(D)

`3/16`

Solution:

`text(Case I )` ` W W B`

`p_1= 3/4 xx 2/4 xx 3/4`

`text(Case II )` ` W B W`

`p_2 = 3/4 xx 2/4 xx 1/4`

`text(Case III )` ` B W W`

`p_3 = 1/4 xx 2/4 xx 1/4`

`:. p = p_1 +p_2 + p_3 = 26/64 =13/32`

Correct Answer is `=>` (A) `13/32`

Q 2581178027

Two coins arc available, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability `3/4`. Given that the outcome is head, the probability that the two-headed coin was chosen, is
WBJEE 2012

(A)

`3/5`

(B)

`2/5`

(C)

`1/5`

(D)

`2/7`

Solution:

Let F denotes fair coin

T denotes two headed

H-denotes head occurs

`:. P(F) = 3/4 , P(T) = 1 - 3/4 = 1/4`

` :. P(T/H) = ( P(T/H) . P(T) )/(P(T/H) . P(T) +P (H/F) . P(F))`

(By Baye's theorem).

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

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

Q 2866034875

In solving any problem, odds against A are `4` to `3` and odds in favour of `B` in solving the same is `7` to `5`. Then, probability that problem will be solved is

(A)

`5//21`

(B)

`16//21`

(C)

`15//84`

(D)

`69//84`

Solution:

Here , `P (A) = 3/7 , P (B) = 7/(12)`,

`P (bar A) = 4/7 , P ( bar B ) = 5/(12)`

Problem will be solved if atleast one

person solves it.

`:.` Required probability

`= 1 - [P (bar A) · P (bar B)]`

` = 1 - 4/7 . 5/(12) = 1 - 5/(21) = (16)/(21)`

Correct Answer is `=>` (B) `16//21`

Q 2105578468

If A and B be mutually exclusive events in a
sample space such that `P(A) = 0.3` and `P(B) = 0.6`, then
`P(bar A cap bar B) = 0.28`

Assertion : If A and B be mutually exclusive events in a sample space such that `P(A) = 0.3` and `P(B) = 0.6`, then `P(bar A cap bar B) = 0.28`

Reason : If A and B are mutually exclusive events, then `P (A cap B) = 0`

(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:

`∵ A` and `B` are mutually exclusive

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

` = P(A) + P(B) - 0`

` = 0.3 + 0.6`

` = 0.9`

`:. P(bar A cap B) = P(bar(A cup B) ) = 1 - P (A cup B)`

` = 1 - 0.9 = 0.1`

Correct Answer is `=>` (D)

Q 2856034874

If A and B are such events that `P(A) > 0` and `P(B) != 1`, then `P(bar A // bar B)` is equal to

(A)

`1 - P(A//B)`

(B)

`1- P(bar A // B)`

(C)

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

(D)

`(P ( bar A))/(P ( bar B))`

Solution:

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

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

Correct Answer is `=>` (C) `(1 - P(A cup B))/(P ( bar B))`

Q 2425391261

The probability that in a year of the `22`nd
century chosen at random, there will be `53`
Sunday, is
UPSEE 2014

(A)

`3/(28)`

(B)

`2/(28)`

(C)

`7/(28)`

(D)

`5/(28)`

Solution:

We know, a leap year is fallen within `4` yr, so its

probability `= (25)/(100) = 1/4`

In a century, the probability of `53` Sunday in a

leap year `= 1/4 xx 2/7 = 2/(28)`

Non-leap year in a century `= 75`

Probability of selecting a non-leap year

` = (75)/(100) xx 3/4`

Probability of `53` Sunday in a non-leap year `= 1/7`

Similarly, in a century, probabilities of `53` Sunday

in a non-leap year `= 3/4 xx 1/7 = 3/(28)`

`:.` Required probability `= 2/(28) + 3/(28) = 5/(28)`

Correct Answer is `=>` (D) `5/(28)`

Q 2407191088

In a college `25%` boys and `10%` girls offer
Mathematics. There are `60%` girls in the
college. If a Mathematics student is chosen at
random, then the probability that the student
is a girl, will be
UPSEE 2009

(A)

`1/6`

(B)

`3/8`

(C)

`5/8`

(D)

`5/6`

Solution:

Let the total number of students be `100`, then in

which `60` girls and `40` boys.

As `25%` of boys offer Mathematics

`= (25)/(100) xx 40 = 10` boys

and `10%` of girls offer Mathematics

`= (10)/(100) xx 60 = 6` girls

`:.` Total number of students, whose offers

Mathematics is `16`.

`:.` Required probability `= 6/(16) = 3/8`

Correct Answer is `=>` (B) `3/8`

Q 2600480318

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.
CBSE-12th 2016

Solution:

Bag `X = 4` white, `2` black.

Bag `Y = 3` white, `3` black

Let A be the event of selecting one white and one black ball.

` E_1 = ` first bag selected

`E_2 =` second bag selected

` P (E_1) = 1/2 P(E_2) = 1/2 `

` P( A//E_1) = 4/6 xx 2/5 + 2/6 xx 4/5 = (16)/(30)`

` P ( A //E_2) = 3/6 xx 3/5 + 3/6 xx 3/5 = (18)/(30)`

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

` P (E_2//A) = ( 1/2 xx (18)/(30) )/( 1/2 xx (16)/(30) + 1/2 xx (18)/(30))`

` = (18)/( 16 + 18) `

` = (18)/(34)`

` = 9/(17)`

Q 2836245172

If the letters of the word 'REGULATION' be arranged at random, the probability that there will be exactly `4` letters between `R` and `E` is

(A)

`1/(10)`

(B)

`1/9`

(C)

`1/5`

(D)

`1/2`

Solution:

There are `10` letters in the word "REGULATION". These `10`

letters can be arranged in `10!` ways.

Exactly `4` letters can be placed between R and E in

`text()^8C_4 xx 4! xx 2! xx 5!` ways.

Hence, required probability ` = ( text()^8C_4 xx 4! xx 2! xx 5!)/(10!) = 1/9`

Correct Answer is `=>` (B) `1/9`

Q 2476378276

If `P(A) = 65, P(B) = 80`, then `P(A cap B)` lies in
the interval
UPSEE 2013

(A)

`[ .30, .80 ]`

(B)

`[.35, .75]`

(C)

`[.4,.70]`

(D)

`[ .45, .65]`

Solution:

`P(A cap B) le min { P(A) , P(B) } =min {.65 , .80 } = 0.65`

`:. P (A cap B) le .65`

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

`ge .65 + .80 -1 = 0.45`

`:. 45 le P (A cap B) le 0.65`

Correct Answer is `=>` (D) `[ .45, .65]`

Q 2816256179

A and B are two events with positive probability, `P(A // B)` is the conditional probability of `A` given `B`, and `bar A` is the complement of `A`.
Consider the following statements
I.` P(A // B) = P(A) => P (B // A) = P(B)`
II. `P(bar B // A) = P (B) => P (B // A) = P(B)`
III. `P(A cap B) = P(A) · P(B) => P (bar A cap bar B) = P(bar A)·P( bar B)`
Which of the above statement(s) is/are correct ?

(A)

I and II

(B)

I and III

(C)

II and III

(D)

I, II and III

Solution:

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

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

`[∵ P(A // B) = P(A)`, given ]

`=> P(A cap B) = P(A) · P(B)`

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

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

Hence, Statement I is correct.

II. `P(bar B // A) = ( P(bar B cap A))/(P(A))`

` => P(bar B) = ( P(A) - P(A cap B))/(P(A))`

`[∵ P(bar B // A)= P(bar B)`, given]

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

`=> P(A cap B) = P(A) · P(B)`

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

` = ( P(A)· P(B))/(P(A)) = P(B)`

Hence, Statement II is correct.

III. Given, `P(A cap B) = P(A) · P(B)`

`=>` A and B are independent events.

`=> bar A` and `bar B` are independent events.

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

Hence, Statement III is also correct.

Correct Answer is `=>` (D) I, II and III

Q 2314234159

The probability that a man will live
another `10` years is `1/4` and the probability
that his wife will live another `10` years is
`1/3` . Then the probability that both will be
alive in `10` years is :
BITSAT Mock

(A)

`(11)/( 12)`

(B)

`7/( 12)`

(C)

`5/( 12)`

(D)

`1/2`

Solution:

Let `P (A) P (B)` be the probability that a

man, woman less live for another `10` yrs.

i.e. `P (A) = 1/4` and `P (B) = 1/3`

`⇒ P (bar A) = 1 − 1/4 = 3/4`

` => P (bar B) = 1 − 1/3 = 2/3`

The probability that both will be alive in

`10` yrs is

` P (bar A, bar B) = P (bar A) P (bar B)`

` = 3/4 . 2/3 = 1/2`

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

Q 1622345231

Probability of solving a particular
question by person `A` is `1/3` and
probability of solving that question by
person `B `is `2/5`. What is the probability
of solving that question by at least one
of them ?
UPSEE 2016

(A)

`2/5`

(B)

`2/3`

(C)

`3/5`

(D)

`7/9`

Solution:

Probability of cycle having no puncture is` (90/100)`

By binomial distribution `P= (9/10), q =1/10`

Required probability ` 5c_5q^0p^5=(9/10)^5`

Correct Answer is `=>` (D) `7/9`

Q 2504434358

If `A` and `B` are two events such that

`P(A cup B) = 3/4 , P(A cap B) =1/4, P(bar (A)) =2/3`, then

`P(bar(A) cap B)` is equal to
UPSEE 2008

(A)

`5/12`

(B)

`3/8`

(C)

`5/8`

(D)

`1/4`

Solution:

for two events `A` and `B`,

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

Since, `P(A cup B) =3/4 , P(A cap B) = 1/4`

and `P(bar(A)) =2/3`

`:. P(A) =1 - P(bar(A)) =1 -2/3 =1/3`

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

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

`=>P(B) =3/4 -1/3 +1/4 = 1-1/2 = 2/3`

Now, `P(bar (A) cap B) = P(B) -P(A cap B)`

`=2/3 -1/4 = (8-3)/12`

`= 5/12`

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

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 1242880733

An urn contains `10` black and `5` white balls . Two balls are drawn from the urn one after the other without replacement . What is the probability that both drawn balls are black ?

(A)

`3/7`

(B)

`4/7`

(C)

`2/7`

(D)

None of theses

Solution:

Let `E` and `F` denote respectively the events that first and second ball drawn are black. we have to find `P(E cap F) ` or ` P(EF)`

Now `P(E)=P(text (black ball in first draw)) = 10/15`

Also given that the first ball drawn is black, i.e., event `E` has occured, now there are `9` black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability pf `F` given that `E` has occured.

i.e `P(FE) = 9/14`

By multiplication cube of probability , we have

`P(E cap F) =P(E) . P(F/E) =10/15 . 9/14 =3/7`

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

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 2886256177

The probability distribution of random variable X with two missing probabilities `p_1` and `p_2` is given below
It is further given that `P(X <= 2) = 0. 25` and `P (X >= 4) = 0. 35`.
Consider the following statements
I. `p_1 = p_2 ` II. `p_1 + p_2 = P(X = 3)`
Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

`∵ P (X <= 2) = 0.25`

`=> P ( X = 1) + P (X = 2) = 0.25`

`=> k + p_1 = 0.25 => p_1 = 0.25 - k`

and `P(X >= 4) = 0.35`

`=> P (X = 4) + P (X = 5) = 0.35`

`=> p_2 + 2k = 0.35`

`=> p_2 = 0.35- 2k => p_1 != p_2`

and `p_1 + p_2 = 0.25 - k + 0.35 - 2k`

`= 0.6 - 3k != P (X = 3)`

Hence, neither I nor II is correct.

Correct Answer is `=>` (D) Neither I nor II

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`

Set - 2

Q 2564734655

Twelve tickets are numbered from `1` to `12`. One ticket is drawn at random, then the probability of
the number to be divisible by `2` or `3`, is
UPSEE 2008

(A)

`2/3`

(B)

`7/12`

(C)

`5/6`

(D)

`3/4`

Solution:

Let `E` be the event of numbers to be divisible by `2`
or `3`.

`:. E = {2, 3, 4, 6, 8, 9, 10, 12}`

`=> n(E) = 8` and `n(S) = 12`

Hence, required probability `= (n(E))/(n(S))`

`= 8/12 = 2/3`

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

Q 2443067843

A card is drawn from a well shuffled pack of
playing cards. The probability that it is a heart
of a king is
UPSEE 2011

(A)

`4/13`

(B)

`15/52`

(C)

`18/52`

(D)

None of these

Solution:

Here, total number of cases `= 52`

and favourables case `= 1`

`:.` Required probability `= 1/52`

Correct Answer is `=>` (D) None of these

Q 2415034860

Odds `8` to `5` against a person who is `40` yr old
living till he is `70` and `4` to `3` against another
person now `50` till he will be living 80.
Probability that one of them will be alive next
`30` yr.
UPSEE 2012

(A)

`59/91`

(B)

`44/91`

(C)

`51/91`

(D)

`32/91`

Solution:

Probability [Person `A` will die in `30` yr] `= 8/(8+5)`

`P(A) = 8/13 =>P(bar (A)) = 5/13`

Similarly, `P(B) = 4/7` or `p(bar(B)) =3/7`

There are two ways in which one person is alive
after `30` yr. `bar(A)B` and `A bar(B)` and event are
independent.
So, required probability

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

`= 5/13 * 4/7 + 8/13 * 3/7 = 44/91`

Correct Answer is `=>` (B) `44/91`

Q 2504434358

If `A` and `B` are two events such that

`P(A cup B) = 3/4 , P(A cap B) =1/4, P(bar (A)) =2/3`, then

`P(bar(A) cap B)` is equal to
UPSEE 2008

(A)

`5/12`

(B)

`3/8`

(C)

`5/8`

(D)

`1/4`

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

Q 2403167948

The probabilities of solving a problem by
three students `A, B` and `C` are `1/2,3/4` and `1/4`
respectively. The probability that the problem
will be solved is
UPSEE 2011

(A)

`3/32`

(B)

`3/16`

(C)

`29/32`

(D)

None of these

Solution:

Given, `P(A) = 1/2 P(B) = 3/4` and `P(C) = 1/4`

`P(bar A) = 1/2., P(B) =1/4` and `P(C) = 3/4`

`:.` Required probability `= 1 - P(bar A cap B cap bar C)`

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

`=1-1/2 xx 1/4 xx3/4`

`=1-3/32 = 29/32`

Correct Answer is `=>` (A) `3/32`

Q 2463167945

If `A` and `B` are two non-empty sets with Be
denoting the complement of set `B` such that
`B subset A`, then which of the following
probability statements hold true?
UPSEE 2011

(A)

`P(A cap B^c)= P(B) -P(A)`

(B)

`P(A cap B^c)= P(A)- P(B)`

(C)

`P(B) le P(A)`

(D)

`P(A) ge P(B)`

Solution:

Since, `B subset A`,

`:. P(A cap B^c)= P(A)- P(B)`

and `P(B) le P(A)`

Correct Answer is `=>` (A)

Q 2826356271

Let `U = { 1,2,3, ... ,20}`. Let A,B and C be the subsets of U. Let A be the set of all numbers, which are perfect squares, B be the set of all numbers, which are multiples of `5` and C be the set of all numbers, which are divisible by `2` and `3`? Consider the following statements
I. A, B and C are mutually exclusive.
II. A, B and C: are mutually exhaustive.
III. The number of elements in the complement set of `A cup B` is `12`.
Which of the above statement(s) is/are correct?

(A)

Both I and II

(B)

Both I and III

(C)

Both II and III

(D)

I, II and III

Solution:

`u = {1, 2,3, ... , 20}`

`A =` Set of all natural numbers which are

perfect square `= { 1, 4, 9, 16}`

`B =` Set of all natural numbers which are

multiples of `5 = {5, 10, 15, 20}`

`C =` Set of all natural numbers which are

divisible by `2` and `3 = {6, 12, 18}`

Here,

`A cup B = {1, 4, 9, 16, 5, 10, 15, 20}`

`=> n (A cup B) = 8`

`=> n (A cup B) = 20 - 8 = 12`

Hence, Statements I and III are correct.

Correct Answer is `=>` (B) Both I and III

Q 2480291117

The probabilities that Mr. A and Mr. B will
die within a year are `1/2` and `1/3` respectively,
then the probability that only one of them
will be alive at the end of the year, is
UPSEE 2015

(A)

`5/6`

(B)

`1/2`

(C)

`2/3`

(D)

None of the above

Solution:

Let A be the event that Mr. A will die and B be the
event that Mr. B will die. Then, required probability

`= P`[(`A` will die and `B` alive) or (`B` will die and `A` alive)]

`= P[(A cap B' ) cup (B cap A')]`

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

[` ·:` these events are mutually exclusive]

`= P(A)* P(B') + P(B)* P(A')`

[`·:A` and `B` are independent]

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

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

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

Q 2612023830

if `(1+4p)/4, (1-p)/4, (1-2p)/4` are probabilities of three mutually exclusive events, then the possible values of `p` belong to the set :

(This question may have multiple correct answers)

(A) `(0, 2/3)`

(B) `(0, 1/2)`

(D) `(-2/3 , 2/3)`

Solution:

We must have, `0 le (1+ 4p)/4 le 1 , 0 le (1-p)/4 le 1` and `0 le (1-2p)/4 le 1`

`=> -1/4 le p le 3/4 , -3 le p le 1 , -3/2 le p le 1/2=> -1/4 le pi le 1/2`

Again the events are mutually exclusive,

so, `0 le (1+ 4p)/4 + (1-p)/4 + (1-2p)/4 le 1`

`=> -3 le p le 1`

Taking intersection of all four intervals of `p`, we get `-1/4 le p le 1/2`

Correct Answer is `=>` (C)

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 2866345275

If `A` and `B` are events such that `P(A cup B) = 0.5, P(bar B) = 0.8` and `P(A // B) = 0.4`, then what is `P(A cap B)` equal to?

(A)

`0.08`

(B)

`0.02`

(C)

`0.8`

(D)

`0.2`

Solution:

Given, `P (A // B) = 0.4, P ( bar B) = 0.8, P (A cup B) = 0.5`

`∵ P(B) + P(bar B) = 1`

`=> P(B) = 1 - P(bar B) = 1 - 0.8 = 0.2`

We know that, multiplication theorem of probability,

`P(A cap B) = P(B) · P(A // B) = 0.2 xx 0.4 = 0.08`

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

Q 2539180912

The probability that A speaks truth is `4/ 5`, while this probability for B is `3//4`. The probability that they contradict each other when asked to speak on a fact, is
BCECE Mains 2015

(A)

`4//5`

(B)

`1//5`

(C)

`7//20`

(D)

`3//20`

Solution:

Consider the following events

`E =` A speaks truth

and `F = B` speaks truth

We have,

`P(E) = 4//5,P(F) = 3//4`

`:.` Required probability `= P(E cap bar F) cup (bar E cap F)`

`= P(E cap bar F) + P(bar E cap F)`

`= P(E) P(bar F) + P(bar E) P(F)`

`= 4//5 xx (1 - 3//4) + (1 - 4//5) xx 3//4`

`= 7//20`

Correct Answer is `=>` (C) `7//20`

Q 2856356274

There are `4` red, `5` blue and `3` green marbles in a basket.
If two marbles are picked randomly, then the probability that both marbles are red is

(A)

`3/7`

(B)

`1/2`

(C)

`1/(11)`

(D)

`1/6`

Solution:

Required probability ` = ( text()^4C_2)/( text()^(12)C_2) = 1/(11)`

Correct Answer is `=>` (C) `1/(11)`

Q 2534156052

If four dice are thrown together. Probability that
the sum of the numbers appearing on them is
`13`, is
UPSEE 2008

(A)

`35/324`

(B)

`5/216`

(C)

`11/216`

(D)

`11/432`

Solution:

Since, `n(S) = 6^4 = 1296`

and permutation that the sum of the numbers
appearing on them is `13`. Total permutation of

`(1, 1, 5, 6) = (4!)/(2!) =12`

Total permutation of `(1, 2, 4, 6) = 4! = 24`

Similarly for `(1, 3, 3, 6) = (4!)/(2!) =12`,

`(1, 2, 5, 5) = 12, (1, 3, 5, 4) = 24, (2, 2, 6, 3) =
12`,

`(2, 2, 5, 4) = 12, (3, 3, 2, 5) =12, (3, 3, 3, 4) = 4`,

`(4, 4, 4, 1) = 4` and `(4, 4, 3, 2)= 12`

`:.` Required probability

`= underline ( 12 + 24 + 12 + 12+ 24 + 12+ 12 + 12 + 4 + 4 + 12)`
`1296`

`= 140/1296 =35/324`

Correct Answer is `=>` (A) `35/324`

Q 2806134978

Two friends `P` and `Q` have equal number of daughters. The two friends have three cinema tickets which are to be distributed among their daughters. If the probability that all the tickets go to daughters of P be `1// 20`, then the number of daughters each has is

(A)

`5`

(B)

`3`

(C)

`4`

(D)

`6`

Solution:

Let each of the friends have `n`

daughters. Then, the probability that all

the tickets go to daughters of `P` is ` (text()^n C_3)/(text()^(2n)C_3)`

` :. (text()^n C_3)/(text()^(2n)C_3) = 1/(20)`

` => ( n(n- 1)(n - 2))/(2n(2n - 1)(2n - 2)) = 1/(20) => (n - 2)/(4(2n - 1)) = 1/(20)`

`=> 5n - 10 = 2n- 1 => 3n = 9`

`:. n = 3`

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

Q 2581178027

(A)

`3/5`

(B)

`2/5`

(C)

`1/5`

(D)

`2/7`

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

Q 2816345279

An observed event `B` can occur after one of the three events `A_1 ,A_2 ,A_3` . If `P(A_1) = P(A_2) = 0.4, P(A_3) = 0. 2`
and `P(B // A_1) = 0.25, P(B // A_2) = 0.4` `P(B // A_3) = 0.125`,
what is the probability of `A_1` after observing `B`?

(A)

`1/3`

(B)

`6/(19)`

(C)

`(20)/(57)`

(D)

`2/5`

Solution:

Required probability `= P ( A_1// B)`

` = ( P(A_1) P (B//A_1))/( ( P(A_1) P (B//A_1) + P(A_2) P (B//A_2) + P(A_3) P (B//A_3))`

` = ( 0.4 xx 0.25)/( 0.4 xx 0.25 + 0.4 xx 0.4 + 0.2 xx 0.125)`

` = (0.1)/( 0.1 - 0.16 + 0.025) = (0.1)/(0.285) = (20)/(57)`

Correct Answer is `=>` (C) `(20)/(57)`

Q 2563112945

A student answers a multiple choice question with `5` alternatives, of which exactly one is correct. The probability that he knows the correct answer is `p, 0 < p < 1`. If he does not know the correct answer, he randomly ticks one answer. Given that he has answered the question correctly, the probability that he did not tick the answer randomly, is
WBJEE 2014

(A)

`(3p)/(4p+3)`

(B)

`(5p)/(3p+2)`

(C)

`(5p)/(4p+1)`

(D)

`(4p)/(3p+1)`

Solution:

Let, `E_1 =` Student does not know the answer

`E_2 =` Student knows the answer

and `E =` Student answer correctly.

`:. P(E_1) =1-p`

`P(E_2)=p`

`P(E/E_2)=1` and `P(E/E_1)=1/5`

`:.` The probability that student did not tick the answer randomly

=The probability that student tick the answer correctly

`=(P(E_2) P(E/E_2))/(P(E_1) P(E/E_1)+ P(E_2) P(E/E_2))`

`=(p(1))/((1-p) 1/5 +p(1))`

`=p/((1-p+5p)/5)= (5p)/(1+4p)`

Correct Answer is `=>` (C) `(5p)/(4p+1)`

Q 2886034877

A contest consists of predicting the results (win, draw or defeat) of `8` matches played by the Indian cricket team. A person sent his entry by predicting at random. The probability that his entry contains `4` correct predictions is

(A)

`(12)/3^8`

(B)

`(70)/3^8`

(C)

`(1120)/3^8`

(D)

`(70)/3^(12)`

Solution:

Here, `n = 8, p = 1/3 , q = 1- p`

`= 1 - 1 = 2/3`

[out of `3` possibilities, only one is correct]

`:.` Required probability `= text()^8C_4 p^4 . q^4`

` = 70 (1/3)^4 (2/3)^4 = (1120)/3^8`

Correct Answer is `=>` (C) `(1120)/3^8`

Q 2439591412

The probability that a student get success in a competition is `3/4` The probability that exactly 2 out of 4 students get success, is
BCECE Stage 1 2014

(A)

`9/41`

(B)

`25/128`

(C)

`1/5`

(D)

`27/128`

Solution:

Given, probability of success, `p = 3/4` and probability of failure, `q =1/4`

Here, `n = 4`

Using binomial distribution

Required probability ` = text()^4C_2 (3/4)^2 (1/4)^2`

`=(4!)/(2 ! 2!) xx 9/16 xx 1/16 =27/128`

Correct Answer is `=>` (D) `27/128`