Engineering
Mathematics
Conditional Probability

Question

Consider a number W = 23 × 3α+1 × 52 × 72 × 11α, α∈N

A set 'S' contains divisor of W. If a divisor is randomly selected from the set S, three events A, B, C are defined as

A : selected divisor is a multiple of 10

B : selected divisor is divisible by 3

C : selected divisor is a multiple of 11

If P(C) = 34 , then answer the following questions.

[Note: P(E) represents probability of event E]

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Linked Question 1

The value of P(BA)+P(CA)  is equal to

35

34

45

3120

Solution

∵    P(C)=34

⇒    P(C)=4×(α+2)×3×3×α4×(α+2)×3×3×(α+1)=34

⇒ 4α = 3α + 3 ⇒ α = 3

(i)  P(BA)+P(CA)=P(A∩B)P(A)+P(C∩A)P(A)

=3×4×2×3×43×5×2×3×4+3×5×2×3×33×5×2×3×4=45+34=3120

(ii) P(A¯∩B∩C)=P(B∩C)−P(A∩B∩C)

=4×4×3×3×34×5×3×3×4−3×4×2×3×34×5×3×3×4=35−310=310.

Linked Question 2

The value of  P(A¯∩B∩C) is equal to

310

89

79

25

Solution

∵P(C)=34

⇒P(C)=4×(α+2)×3×3×α4×(α+2)×3×3×(α+1)=34

⇒ 4α = 3α + 3 ⇒ α = 3

(i)  P(BA)+P(CA)=P(A∩B)P(A)+P(C∩A)P(A)

=3×4×2×3×43×5×2×3×4+3×5×2×3×33×5×2×3×4=45+34=3120

(ii)  P(A¯∩B∩C)=P(B∩C)−P(A∩B∩C)

=4×4×3×3×34×5×3×3×4−3×4×2×3×34×5×3×3×4=35−310=310.