CONDITIONAL PROBABILITY PROBLEMS WITH SOLUTIONS

Problem 1 :

A problem in Mathematics is given to three students whose chances of solving it are 1/3, 1/4 and 1/5 (i) What is the probability that the problem is solved? (ii) What is the probability that exactly one of them will solve it?

Solution :

Let "A", "B" and "C" be the events of solving problems by each students respectively.

P(A)  =  1/3, P(B)  =  1/4 and P(C)  =  1/5

(i) What is the probability that the problem is solved? 

P(Problem solved)  =  P(At least one solving)

  =  1 - P(None solving the problem)

=  1 - P(A' n B' n C')

=  1 -  P(A') ⋅ P(B') ⋅  P(C')

  =  1 -  (2/3) (3/4) (4/5)

  =  1 - (2/5)

  =  (5 - 2) / 5

P(Problem solved)  =  3/5

(ii) What is the probability that exactly one of them will solve it

P(exactly one of them will solve it) 

  =  P(A' n B' n c) + P(A' n B n c') + P(A n B' n c')

  =  P(A') P(B') P(C) +  P(A') P(B) P(C') +  P(A) P(B') P(C')

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

  =  (6/60) + (8/60) + (12/60)

  =  (6 + 8 + 12)/60

  =  26/60

P(exactly one of them will solve it)  =  13/30

Problem 2 :

The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15.

(i)  If the oil had to be changed, what is the probability that a new oil filter is needed?

(ii)  If a new oil filter is needed, what is the probability that the oil has to be changed?

Solution :

Let "A" and "B" the event of changing oil and new oil filter respectively.

P(A)  =  0.30, P(B)  =  0.40, P(AnB)  =  0.15

(i)  If the oil had to be changed, what is the probability that a new oil filter is needed?

Here we have to find the probability that a new oil filter is needed, if the oil had to be changed.

The event B depends on A.

P(B/A)  =  P(AnB)/P(A)

  =  0.15 / 0.30

  =  1/2

P(B/A)  =  0.5

(ii)  If a new oil filter is needed, what is the probability that the oil has to be changed?

The event A depends on B.

P(A/B)  =  P(AnB)/P(B)

  =  0.15 / 0.40

  =  3/8

P(A/B)  =  0.375

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