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Question: A candidate takes three tests in succession and the probability of passing the first test is \(P\). ...

A candidate takes three tests in succession and the probability of passing the first test is PP. The probability of passing each succeeding test is PP or P2\dfrac{P}{2} according as he passes or fails in the preceding one. The candidate is selected if he passes at least two tests. The probability that the candidate is selected is
A. P2(2−P){P^2}\left( {2 - P} \right)
B. P(2−P)P\left( {2 - P} \right)
C. P+P2+P3P + {P^2} + {P^3}
D. P2(1−P){P^2}\left( {1 - P} \right)

Explanation

Solution

First make the cases where as least two papers are passed. Then calculate the probability of each case. Next, add all the probabilities of the formed cases. Solve the expression further by taking terms common and simplify the expression to get the required answer.

Complete step by step answer:

We are given that the candidate takes 3 tests in a succession and the candidate is selected only when he passes at least two tests.
It is given that the probability of passing the test after each success is PP and after each failure is P2\dfrac{P}{2}
The probability of failing a test is equal to 1−P1 - P because the sum of probability of an event and probability of not an event is 1.
Now, suppose the candidate passes all the three tests
Then the probability of passing all the three tests is P.P.P = P3P.P.P{\text{ }} = {\text{ }}{P^3}
Let us now let that the candidate passes the first two test and fails in the last test.
Then the probability of the candidate passing the first two tests and fails in third test is P.P.(1−P)P.P.\left( {1 - P} \right)
Now, take the case where the candidate passes the first and the third test and fails in the second test.
Also, the probability of the successive test after a failure be P2\dfrac{P}{2}
Then the probability will be P.(1−P)P2P.\left( {1 - P} \right)\dfrac{P}{2}
But, if the candidate fails in the first test and passes other two tests, then the probability is (1−P)P2(P)\left( {1 - P} \right)\dfrac{P}{2}\left( P \right)
the required probability is the sum of the probabilities of all the cases of selection of a candidate.
That is P3+P.P.(1−P)+P.(1−P)P2+(1−P)P2(P){P^3} + P.P.\left( {1 - P} \right) + P.\left( {1 - P} \right)\dfrac{P}{2} + \left( {1 - P} \right)\dfrac{P}{2}\left( P \right)
On simplifying the given expression, we get,
P(P2+P.(1−P)+(1−P)P2+(1−P)P2)P\left( {{P^2} + P.\left( {1 - P} \right) + \left( {1 - P} \right)\dfrac{P}{2} + \left( {1 - P} \right)\dfrac{P}{2}} \right)
Now, take 1−P1 - P common and solve
P(P2+(1−P)(P+P2+P2)) ⇒P(P2+(1−P)(P+P)) ⇒P(P2+(1−P)(2P))  P\left( {{P^2} + \left( {1 - P} \right)\left( {P + \dfrac{P}{2} + \dfrac{P}{2}} \right)} \right) \\\ \Rightarrow P\left( {{P^2} + \left( {1 - P} \right)\left( {P + P} \right)} \right) \\\ \Rightarrow P\left( {{P^2} + \left( {1 - P} \right)\left( {2P} \right)} \right) \\\
Now solve the inner bracket
P(P2+(1−P)(2P)) ⇒P(P2+2P−2P2) ⇒P(2P−P2)  P\left( {{P^2} + \left( {1 - P} \right)\left( {2P} \right)} \right) \\\ \Rightarrow P\left( {{P^2} + 2P - 2{P^2}} \right) \\\ \Rightarrow P\left( {2P - {P^2}} \right) \\\
Again, taking PP common to get,
⇒P2(2−P)\Rightarrow {P^2}\left( {2 - P} \right)
Hence, option A is correct.

Note: The probability of the failing a test is not given directly in the question. Students have to use the properties of probability to find the probability of failing a test when probability of passing a test is given. The sum of probability of an event and probability of not an event is equal to 1.