Question

6. Kevin will soon be taking exams in math, physics, and French. He estimates the probabilities of his passing these exams to be as follows:

 Math: 0.9
 Physics: 0.8
 French: 0.7
Kevin is willing to assume that the results of the three exams are independent of each other. Find the probability of each event.
a. Kevin will pass all three exams.
b. Kevin will pass math but fail the other two exams.
c. Kevin will pass exactly one of the three exams.

Answer 1

Answer: The required probabilities is

(a) 0.504 (b) 0.054 (c) 0.092.

Step-by-step explanation: Given that Kevin will soon be taking exams in math, physics, and French. He estimates the probabilities of his passing these exams to be as follows:

`P(m)=0.9,~~~P(p)=0.8,~~~P(f)=0.7.`

The results of these three exams are independent of each other.

We know that if A and B are independent events, then

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

(a) The probability that Kevin pass all three exams is

`P(m\cap p\cap f)=P(m)* P(p)* P(f)=0.9*0.8*0.7=0.504.`

(b) The probability that Kevin pass math but fail the other exams is

`P(m\cap \bar{p}\cap \bar{f})=P(m)P(\bar{p})P(\bar{f})=0.9(1-0.8)(1-0.7)=0.9*0.2*0.3=0.054.`

(c) The probability that Kevin pass exactly one of the three exams is

`P(m\cap \bar{p}\cap \bar{f})+P(\bar{m}\cap p\cap \bar{f})+P(\bar{m}\cap \bar{p}\cap f)\\\\=0.9(1-0.8)(1-0.7)+(1-0.9)0.8(1-0.7)+(1-0.9)(1-0.8)0.7\\\\=0.054+0.024+0.014\\\\=0.092.`

Thus, the required probabilities is

(a) 0.504 (b) 0.054 (c) 0.092.