`star` Axiomatic Approach to Probability

`star` Probability of an event

`star` Probabilities of equally likely outcomes

`star` Probability of the event `"‘A or B’"`

`star` Probability of event `"‘not A’"`

`star` Probability of an event

`star` Probabilities of equally likely outcomes

`star` Probability of the event `"‘A or B’"`

`star` Probability of event `"‘not A’"`

`\color{green} ✍️` we have studied some methods of assigning probability to an event associated with an experiment having known the number of total outcomes.

`\color{green} ✍️ ` Axiomatic approach is another way of describing probability of an event. In this approach some axioms or rules are depicted to assign probabilities.

Let `S` be the sample space of a random experiment.

The probability `P` is a real valued function whose domain is the power set of `S` and range is the interval `[0,1]` satisfying the following axioms

`color(red)((i))` For any event `E, \ \ P (E) ≥ 0`

`color(red)((ii))\ \ P (S) = 1`

`color(red)((iii))` If `E` and `F` are `color(blue)("mutually exclusive events")`, then `color(green)(P(E ∪ F) = P(E) + P(F).)`

It follows from `(iii)` that `P(φ) = 0.`

To prove this, we take `F = φ` and note that `E` and `φ` are disjoint events.

Therefore, from axiom `(iii),` we get `color(blue)(P (E ∪ φ) = P (E) + P (φ))` or `color(green)(P(E) = P(E) + P (φ))` i.e. `color(red)(P (φ) = 0.)`

Let `S` be a sample space containing outcomes `ω_1 , ω_2 ,...,ω_n` , i.e.,

`S = {ω_1, ω_2, ..., ω_n}`

It follows from the axiomatic definition of probability that

`color(red)((i)) color(green)(0 ≤ P (ω_i) ≤ 1)` for each `ω_i ∈ S`

`color(red)((ii)) P (ω_1) + P (ω_2) + ... + P (ω_n) = 1`

`color(red)((iii))` For any event `A,` `color(green)(P(A) = Σ P(ω_i ), \ \ ω_i ∈ A.)`

For example, in ‘a coin tossing’ experiment we can assign the number `1/2 ` to each of the outcomes H and T.

`\color{green} ✍️ ` Axiomatic approach is another way of describing probability of an event. In this approach some axioms or rules are depicted to assign probabilities.

Let `S` be the sample space of a random experiment.

The probability `P` is a real valued function whose domain is the power set of `S` and range is the interval `[0,1]` satisfying the following axioms

`color(red)((i))` For any event `E, \ \ P (E) ≥ 0`

`color(red)((ii))\ \ P (S) = 1`

`color(red)((iii))` If `E` and `F` are `color(blue)("mutually exclusive events")`, then `color(green)(P(E ∪ F) = P(E) + P(F).)`

It follows from `(iii)` that `P(φ) = 0.`

To prove this, we take `F = φ` and note that `E` and `φ` are disjoint events.

Therefore, from axiom `(iii),` we get `color(blue)(P (E ∪ φ) = P (E) + P (φ))` or `color(green)(P(E) = P(E) + P (φ))` i.e. `color(red)(P (φ) = 0.)`

Let `S` be a sample space containing outcomes `ω_1 , ω_2 ,...,ω_n` , i.e.,

`S = {ω_1, ω_2, ..., ω_n}`

It follows from the axiomatic definition of probability that

`color(red)((i)) color(green)(0 ≤ P (ω_i) ≤ 1)` for each `ω_i ∈ S`

`color(red)((ii)) P (ω_1) + P (ω_2) + ... + P (ω_n) = 1`

`color(red)((iii))` For any event `A,` `color(green)(P(A) = Σ P(ω_i ), \ \ ω_i ∈ A.)`

`color{blue} "Key Point : "` It may be noted that the singleton `{ω_i}` is called elementary event and for notational convenience, we write `P(ω_i ) ` for `P({ω_i }`).

For example, in ‘a coin tossing’ experiment we can assign the number `1/2 ` to each of the outcomes H and T.

Q 3110191910

Let a sample space be `S = {ω1, ω2,..., ω6}.`Which of the following assignments of probabilities to each outcome are valid?

`(a) 1/6 \ \ \ 1/6 \ \ \ 1/6 \ \ \1/6 \ \ \ \ \1/6 \ \ \ \1/6`

`(b) 1\ \ \ \ \ 0 \ \ \ \0 \ \ \ \ \0 \ \ \ \ \0 \ \ \ \ \0`

`(c)1/8 \ \ \ \ 2/3\ \ \ 1/3\ \ \ \ 1/3\ \ \ \- 1/4\ \ \ \- 1/3`

`(d) 1/12\ \ \ \1/12\ \ \ \ 1/6\ \ \ 1/6 \ \ \ \1/6\ \ \ \ \3/2`

`(e) 0.1 \ \ \ \0.2 \ \ \ \ 0.3\ \ \ \0.4\ \ \ \ 0.5\ \ \ \ \ 0.6`

`(a) 1/6 \ \ \ 1/6 \ \ \ 1/6 \ \ \1/6 \ \ \ \ \1/6 \ \ \ \1/6`

`(b) 1\ \ \ \ \ 0 \ \ \ \0 \ \ \ \ \0 \ \ \ \ \0 \ \ \ \ \0`

`(c)1/8 \ \ \ \ 2/3\ \ \ 1/3\ \ \ \ 1/3\ \ \ \- 1/4\ \ \ \- 1/3`

`(d) 1/12\ \ \ \1/12\ \ \ \ 1/6\ \ \ 1/6 \ \ \ \1/6\ \ \ \ \3/2`

`(e) 0.1 \ \ \ \0.2 \ \ \ \ 0.3\ \ \ \0.4\ \ \ \ 0.5\ \ \ \ \ 0.6`

(a) Condition (i): Each of the number `p(ωi)` is positive and less than one.

Condition (ii): Sum of probabilities

`= 1/6+1/6+1/6+1/6+1/6+1/6=1`

Therefore, the assignment is valid

(b) Condition (i): Each of the number p(ωi) is either 0 or 1.

Condition (ii) Sum of the probabilities `= 1 + 0 + 0 + 0 + 0 + 0 = 1`

Therefore, the assignment is valid

(c) Condition (i) Two of the probabilities `p(ω5)` and `p(ω6)` are negative, the assignment

is not valid

(d) Since `p(omega_6) = -3/2>1` the assignment is not valid

(e) Since, sum of probabilities `= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1,` the assignment

is not valid.

`\color{fuchsia} {ul★ " Probability of an event"}`

Let `S` be a sample space associated with the experiment ‘examining three consecutive pens produced by a machine and classified as Good (non-defective) and bad (defective)’.

We may get 0, 1, 2 or 3 defective pens as result of this examination.

A sample space associated with this experiment is

` " " color(blue)(S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG})`

where `B` stands for a defective or bad pen and `G` for a non – defective or good pen.

Let the probabilities assigned to the outcomes be as follows

`tt ( (text{Sample point :} , BBB , BBG , BGB , GBB , BGG , GBG , GGB , GGG) , ( text{Probability:} , 1/8 , 1/8 , 1/8 , 1/8 , 1/8 , 1/8 , 1/8 , 1/8))`

Let `color(red)("event A: there is exactly one defective pen")` and

`color(red)("event B: there are atleast two defective pens.")`

Hence `A = {BGG, GBG, GGB}` and `B = {BBG, BGB, GBB, BBB}`

Now `color(red)(P(A)) = Sigma P(omega_i) AA omega_i in A`

`" " = P (BGG) +P(GBG)+P(GGB) = 1/8+1/8+1/8 = 3/8`

and `color(red)(P(B)) = Sigma P(omega_i) AA omega_i in B`

`" " = P(BBG) + P(BGB) + P(GBB) + P(BBB) = 1/8+1/8+1/8+1/8 = 4/8 = 1/2`

Let `S` be a sample space associated with the experiment ‘examining three consecutive pens produced by a machine and classified as Good (non-defective) and bad (defective)’.

We may get 0, 1, 2 or 3 defective pens as result of this examination.

A sample space associated with this experiment is

` " " color(blue)(S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG})`

where `B` stands for a defective or bad pen and `G` for a non – defective or good pen.

Let the probabilities assigned to the outcomes be as follows

`tt ( (text{Sample point :} , BBB , BBG , BGB , GBB , BGG , GBG , GGB , GGG) , ( text{Probability:} , 1/8 , 1/8 , 1/8 , 1/8 , 1/8 , 1/8 , 1/8 , 1/8))`

Let `color(red)("event A: there is exactly one defective pen")` and

`color(red)("event B: there are atleast two defective pens.")`

Hence `A = {BGG, GBG, GGB}` and `B = {BBG, BGB, GBB, BBB}`

Now `color(red)(P(A)) = Sigma P(omega_i) AA omega_i in A`

`" " = P (BGG) +P(GBG)+P(GGB) = 1/8+1/8+1/8 = 3/8`

and `color(red)(P(B)) = Sigma P(omega_i) AA omega_i in B`

`" " = P(BBG) + P(BGB) + P(GBB) + P(BBB) = 1/8+1/8+1/8+1/8 = 4/8 = 1/2`

Let a sample space of an experiment be

`S = {ω_1, ω_2,..., ω_n}`.

Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same

i.e. `P ( ω_i ) = p` , for all ` ω_i ∈ S` where ` 0 le p le 1`

Since `sum_(i=1)^n P( ω_i) =1` i.e, `p + p + ... + p (n" times") = 1 `

`np = 1` i.e, `p = 1/n`

Let `S` be a sample space and E be an event, such that `n(S) = n` and `n(E) = m`.

If each out come is equally likely, then it follows that

`color(red)(P(E) = m/n = ( text (Number of outcomes favourable to E ) )/( text ( Total possible outcomes ) ) )`

`S = {ω_1, ω_2,..., ω_n}`.

Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same

i.e. `P ( ω_i ) = p` , for all ` ω_i ∈ S` where ` 0 le p le 1`

Since `sum_(i=1)^n P( ω_i) =1` i.e, `p + p + ... + p (n" times") = 1 `

`np = 1` i.e, `p = 1/n`

Let `S` be a sample space and E be an event, such that `n(S) = n` and `n(E) = m`.

If each out come is equally likely, then it follows that

`color(red)(P(E) = m/n = ( text (Number of outcomes favourable to E ) )/( text ( Total possible outcomes ) ) )`

Let us now find the probability of event `"‘A or B’"`, i.e., `P (A ∪ B)`

Let `A = {HHT, HTH, THH}` and `B = {HTH, THH, HHH} ` be two events associated with ‘tossing of a coin thrice’

Clearly `A ∪ B = {HHT, HTH, THH, HHH}`

Now `P (A ∪ B) = P(HHT) + P(HTH) + P(THH) + P(HHH)`

If all the outcomes are equally likely, then

`P (A ∪ B) = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2`

Also` P(A) = P(HHT) + P(HTH) + P(THH) = 3/8`

and `P(B) = P(HTH) + P(THH) + P(HHH) = 3/8`

Therefore` P(A) + P(B) = 3/8 + 3/8 = 6/8`

It is clear that` P(A∪ B) ≠ P(A) + P(B)`

The points `HTH` and `THH ` are common to both `A` and `B .`

In the computation of `P(A) + P(B)` the probabilities of points `HTH` and `THH`, i.e., the elements of `A ∩B` are included twice.

Thus to get the probability `P(A ∪ B) `we have to subtract the probabilities of the sample points in `A ∩ B` from `P(A) + P(B)`

i.e. `P(A∪B) = P(A) + P(B) − ΣP(ω_i ),∀ ω_i ∈ A∩ B`

`= P(A) + P(B) − P(A ∩B)`

Thus we observe that, `P(A∪ B) = P(A) + P(B) − P(A ∩B)`

In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have

` P(A ∪ B ) = Σ p(ω_i ),∀ ω_i ∈ A ∪ B`.

Since `A∪B = (A–B) ∪ (A ∩B)∪(B–A)` ,

we have

`P(A ∪ B) = [ Σ P(ω_i ) ∀ ω_i ∈ (A–B)] + [ Σ P(ω_i ) ∀ ω_i ∈ A ∩ B ] + [ ΣP ( ω_i ) ∀ ω_i ∈ B – A]`

(because `A–B, A ∩ B` and `B – A `are mutually exclusive) ... (1)

Also `P(A) + P(B) = [Σ p(ω_i ) ∀ω_i ∈A]+[ Σ p(ω_i ) ∀ω_i ∈B]`

`= [ Σ P (ω_i )∀ω_i ∈(A–B)∪(A∩B)] + [ Σ P (ω_i )∀ω_ i ∈(B – A)∪(A∩B)]`

`= [ Σ P (ω_i )∀ω_i ∈(A – B ) ] + [ Σ P ( ω_i ) ∀ ω_i ∈(A∩B)] + [ Σ P (ω_i ) ∀ω_i ∈ (B–A)] +

[ Σ P(ω_i )∀ω_ i ∈(A∩ B)]`

`= P(A∪B) + [ Σ P (ω_i )∀ ω_i ∈A∩B]` [using (1)]

`= P(A∪ B)+P(A ∩B) `.

Hence `P(A∪B)=P (A)+P(B) – P(A∩B)` .

Alternatively, it can also be proved as follows:

`A ∪ B = A ∪ (B – A)`, where A and `B – A` are mutually exclusive,

and `B = (A ∩ B) ∪ (B – A)`, where `A ∩ B` and B – A are mutually exclsuive.

Using Axiom (iii) of probability, we get

`P (A ∪B) = P (A) + P (B – A)` ... (2)

and `P(B) = P ( A ∩ B) + P (B – A)` ... (3)

Subtracting (3) from (2) gives

`P (A ∪ B) – P(B) = P(A) – P (A ∩ B)`

or `P(A ∪ B) = P(A) + P (B) – P (A ∩ B)`

The above result can further be verified by observing the Venn Diagram (Fig 16.1)

`color(red)(=> P(A ∪B) = P(A) + P(B)) `,

`color(blue)("If A and B are disjoint sets")`, i.e., they are mutually exclusive events, then `color(red)(A ∩ B = φ)`

Therefore `color(blue)(P(A∩B) = P (φ) = 0)`

Thus, for mutually exclusive events `A` and `B`, we have

`P(A ∪B) = P(A) + P(B) `,

which is Axiom (iii) of probability.

Let `A = {HHT, HTH, THH}` and `B = {HTH, THH, HHH} ` be two events associated with ‘tossing of a coin thrice’

Clearly `A ∪ B = {HHT, HTH, THH, HHH}`

Now `P (A ∪ B) = P(HHT) + P(HTH) + P(THH) + P(HHH)`

If all the outcomes are equally likely, then

`P (A ∪ B) = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2`

Also` P(A) = P(HHT) + P(HTH) + P(THH) = 3/8`

and `P(B) = P(HTH) + P(THH) + P(HHH) = 3/8`

Therefore` P(A) + P(B) = 3/8 + 3/8 = 6/8`

It is clear that` P(A∪ B) ≠ P(A) + P(B)`

The points `HTH` and `THH ` are common to both `A` and `B .`

In the computation of `P(A) + P(B)` the probabilities of points `HTH` and `THH`, i.e., the elements of `A ∩B` are included twice.

Thus to get the probability `P(A ∪ B) `we have to subtract the probabilities of the sample points in `A ∩ B` from `P(A) + P(B)`

i.e. `P(A∪B) = P(A) + P(B) − ΣP(ω_i ),∀ ω_i ∈ A∩ B`

`= P(A) + P(B) − P(A ∩B)`

Thus we observe that, `P(A∪ B) = P(A) + P(B) − P(A ∩B)`

In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have

` P(A ∪ B ) = Σ p(ω_i ),∀ ω_i ∈ A ∪ B`.

Since `A∪B = (A–B) ∪ (A ∩B)∪(B–A)` ,

we have

`P(A ∪ B) = [ Σ P(ω_i ) ∀ ω_i ∈ (A–B)] + [ Σ P(ω_i ) ∀ ω_i ∈ A ∩ B ] + [ ΣP ( ω_i ) ∀ ω_i ∈ B – A]`

(because `A–B, A ∩ B` and `B – A `are mutually exclusive) ... (1)

Also `P(A) + P(B) = [Σ p(ω_i ) ∀ω_i ∈A]+[ Σ p(ω_i ) ∀ω_i ∈B]`

`= [ Σ P (ω_i )∀ω_i ∈(A–B)∪(A∩B)] + [ Σ P (ω_i )∀ω_ i ∈(B – A)∪(A∩B)]`

`= [ Σ P (ω_i )∀ω_i ∈(A – B ) ] + [ Σ P ( ω_i ) ∀ ω_i ∈(A∩B)] + [ Σ P (ω_i ) ∀ω_i ∈ (B–A)] +

[ Σ P(ω_i )∀ω_ i ∈(A∩ B)]`

`= P(A∪B) + [ Σ P (ω_i )∀ ω_i ∈A∩B]` [using (1)]

`= P(A∪ B)+P(A ∩B) `.

Hence `P(A∪B)=P (A)+P(B) – P(A∩B)` .

Alternatively, it can also be proved as follows:

`A ∪ B = A ∪ (B – A)`, where A and `B – A` are mutually exclusive,

and `B = (A ∩ B) ∪ (B – A)`, where `A ∩ B` and B – A are mutually exclsuive.

Using Axiom (iii) of probability, we get

`P (A ∪B) = P (A) + P (B – A)` ... (2)

and `P(B) = P ( A ∩ B) + P (B – A)` ... (3)

Subtracting (3) from (2) gives

`P (A ∪ B) – P(B) = P(A) – P (A ∩ B)`

or `P(A ∪ B) = P(A) + P (B) – P (A ∩ B)`

The above result can further be verified by observing the Venn Diagram (Fig 16.1)

`color(red)(=> P(A ∪B) = P(A) + P(B)) `,

`color(blue)("If A and B are disjoint sets")`, i.e., they are mutually exclusive events, then `color(red)(A ∩ B = φ)`

Therefore `color(blue)(P(A∩B) = P (φ) = 0)`

Thus, for mutually exclusive events `A` and `B`, we have

`P(A ∪B) = P(A) + P(B) `,

which is Axiom (iii) of probability.

Also, we know that `A′` and `A` are mutually exclusive and exhaustive events i.e.,

`A ∩ A′ = φ` and `A ∪ A′ = S`

or `color(blue)(P(A ∪ A′) = P(S))`

Now `P(A) + P(A′) = 1`,

or `color(red)(P( A′ ) = P ( "not " A ) = 1 – P(A))`

`color(red)(=> "Consider the event " A = {2, 4, 6, 8})`

associated with the experiment of drawing a card from a deck of ten cards numbered from `1` to `10.`

Clearly the `color(blue)("sample space is "S = {1, 2, 3, ...,10})`

If all the outcomes `1, 2, ...,10` are considered to be equally likely, then the probability of each outcome is `1/10`

Now `P(A) = P(2) + P(4) + P(6) + P(8)`

`" " =1/10 + 1/10 + 1/10 + 1/10 = 4/10 = 2/5`

Also event ‘not `A’ = A′ = {1, 3, 5, 7, 9, 10}`

Now `color(blue)(P(A′)) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10)`

` " " = 6/10 = 3/5`

Thus , `color(red)(P(A') )= 3/5 = 1-2/5 = color(red)(1 -P(A))`

`A ∩ A′ = φ` and `A ∪ A′ = S`

or `color(blue)(P(A ∪ A′) = P(S))`

Now `P(A) + P(A′) = 1`,

or `color(red)(P( A′ ) = P ( "not " A ) = 1 – P(A))`

`color(red)(=> "Consider the event " A = {2, 4, 6, 8})`

associated with the experiment of drawing a card from a deck of ten cards numbered from `1` to `10.`

Clearly the `color(blue)("sample space is "S = {1, 2, 3, ...,10})`

If all the outcomes `1, 2, ...,10` are considered to be equally likely, then the probability of each outcome is `1/10`

Now `P(A) = P(2) + P(4) + P(6) + P(8)`

`" " =1/10 + 1/10 + 1/10 + 1/10 = 4/10 = 2/5`

Also event ‘not `A’ = A′ = {1, 3, 5, 7, 9, 10}`

Now `color(blue)(P(A′)) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10)`

` " " = 6/10 = 3/5`

Thus , `color(red)(P(A') )= 3/5 = 1-2/5 = color(red)(1 -P(A))`

Q 3130191912

One card is drawn from a well shuffled deck of 52 cards. If each outcome

is equally likely, calculate the probability that the card will be

(i) a diamond

(ii) not an ace

(iii) a black card (i.e., a club or, a spade)

(iv) not a diamond

(v) not a black card.

is equally likely, calculate the probability that the card will be

(i) a diamond

(ii) not an ace

(iii) a black card (i.e., a club or, a spade)

(iv) not a diamond

(v) not a black card.

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.

(i) Let A be the event 'the card drawn is a diamond'

Clearly the number of elements in set A is 13.

Therefore `P(A) = 13/52=1/4`

i.e. Probability of a diamond card = -

(ii) We assume that the event ‘Card drawn is an ace’ is B

Therefore ‘Card drawn is not an ace’ should be B′.

We know that `P(B')= 1- P(B)=1-4/52+1- 1/14 = 12/13`

(iii) Let C denote the event ‘card drawn is black card’

Therefore, number of elements in the set C = 26

`i.e P(C) = 26/52 = 1/2`

Thus, Probability of a black card `= 1/2`

(iv) We assumed in (i) above that A is the event ‘card drawn is a diamond’,

so the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’

Now `P("not A") = 1 – P(A) = 1- 1/4 = 3/4`

(v) The event ‘card drawn is not a black card’ may be denoted as C′ or ‘not C’.

We know that `P("not C") = 1 - P(C) = 1 - 1/2 = 1/2`

Therefore, Probability of not a black card `= 1/2`

Q 3160191915

A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be (i) red, (ii) yellow, (iii) blue, (iv) not blue, (v) either red or yellow.

There are 9 discs in all so the total number of possible outcomes is 9.

Let the events A, B, C be defined as

A: ‘the disc drawn is red’

B: ‘the disc drawn is yellow’

C: ‘the disc drawn is blue’.

(i) The number of red discs `= 4, i.e., n (A) = 4`

Hence `P(A) = 4/9`

(ii) The number of yelow discs `= 2, i.e., n (B) = 2`

Therefore , `P(B) = 2/9`

(iii) The number of blue discs `= 3, i.e., n(C) = 3`

Therefore `P(C) = 3/9 = 1/3`

(iv) Clearly the event ‘not blue’ is ‘not C’. We know that P(not C) = 1 – P(C)

(V) The event ‘either red or yellow’ may be described by the set ‘A or C’ Since, A and C are mutually exclusive events, we have

`P(A or C) = P (A ∪ C) = P(A) + P(C) = 4/9 +1/3 + 7/9`

Q 3170191916

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that

(a) Both Anil and Ashima will not qualify the examination.

(b) Atleast one of them will not qualify the examination and

(c) Only one of them will qualify the examination.

(a) Both Anil and Ashima will not qualify the examination.

(b) Atleast one of them will not qualify the examination and

(c) Only one of them will qualify the examination.

Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. Given that

`P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02.`

Then

(a) The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as `E´ ∩ F´.`

Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e.,

Ashima will not qualify the examination.

Also `E´ ∩ F´ = (E ∪ F)´` (by Demorgan's Law)

Now `P(E ∪ F) = P(E) + P(F) – P(E ∩ F)`

or `P(E ∪ F) = 0.05 + 0.10 – 0.02 = 0.13`

Therefore `P(E´ ∩ F´) = P(E ∪ F)´ = 1 – P(E ∪ F) = 1 – 0.13 = 0.87`

(b) P (atleast one of them will not qualify)

`= 1 – P` (both of them will qualify)

`= 1 – 0.02 = 0.98`

(c) The event only one of them will qualify the examination is same as the event

either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima

will qualify) i.e., `E ∩ F´ or E´ ∩ F,` where `E ∩ F`´ and `E´ ∩ F` are mutually exclusive.

Therefore, P(only one of them will qualify) = P(E ∩ F´ or E´ ∩ F)

`= P(E ∩ F´) + P(E´ ∩ F) = P (E) – P(E ∩ F) + P(F) – P (E ∩ F)`

`= 0.05 – 0.02 + 0.10 – 0.02 = 0.11`

Q 3131101022

A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man? (b) one man? (c) two men?

The total number of persons `= 2 + 2 = 4.` Out of these four person, two can be selected in `text()^4C_2`

(a) No men in the committee of two means there will be two women in the committee. Out of two women, two can be selected in `text()^2C_2 =1` way.

Therefore P(on man) `= (text()^2C_2)/(text()^4C_2) = 1xx2xx1/(4xx3) = 1/6`

(b) One man in the committee means that there is one woman. One man out of 2

can be selected in `text()^2C_1` ways and one woman out of 2 can be selected in `text()^2C_1` ways

Together they can be selected in `text()^2C_1 xx text()^2C_1` ways.

Therefore P(One man) `= text()^2C_1 xx (text()^2C_1)/(text()^4C_2) = 2/3`

(c) Two men can be selected in `text()^2C_2` way

Hence P(Two men) `= (text()^2C_2)/(text()^4C_1) = 1/(text()^4C_2)= 1/6`