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Kirti and Justin are both preparing to take a driving test. They each learned to drive separately, so the results of the tests are independent. The probability that Kirti passes is 0.6 and the probability that Justin passes is 0.4.
Calculate the probability that:

a) both pass the test
b) Neither passes the test
c) Kirti passes the test, but Justin doesn't pass
e) exactly one of Kirti and Justin passes.​

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rvkacademic

Answer:

a) 0.24

b) 0.24

c) 0.36

d) 0.52

Step-by-step explanation:

Let K represent the event that Kirti passes the test and J represent the event that Justin passes the test.

Given

P(K) = 0.6 we can compute P(K') = 1-0.6 = 0.4 where K' represents the complement of the event K i.e. Kirti does not pass the test

In a similar manner P(J) = 0.4 and P(J') = 1-0.4 = 0.6

a) P(both pass the test) = P(K and J) which in set notation is

P(K∩J) = P(K) x P(J) since the events are independent

So P(K∩J) = 0.6 x 0.4 = 0.24   (Answer a)

b) P(neither passing test) = P(K not passing).P(J not passing) = P(K' ∩ J') = (1-0.6) x (1-0.4) = 0.4 x 0.6 = 0.24   (Answer b)

c) P(Kirti passing test but Justin not passing) =  P(K).P(J') = (1-0.6) x (1-0.4) = 0.6 x 0.6 = 0.36   (Answer c)

d) P(exactly one of Kirti and Justin passing) = P(K pass and Justin not pass) + P(K not pass and Justin pass) which in set notation can be expressed as

P(K∩J') + P(K'∩J)

Noting that K and J are independent events

P(K∩J') =P(K).P(J') = 0.6 x 0.6 = 0.36

P(K'∩J)  = P(K').P(J) = 0.4 x 0.4 = 0.16

So P(K∩J') + P(K'∩J)  = 0.36 + 0.16 = 0.52   (Answer d)

Note

For part d we can also compute using the formula for two independent events

P(K or J but not both) = P(K) + P(J) - 2.P(K∩J)

= 0.6 + 0.4 - 2(0.24) = 1 - 0.48 = 0.52