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# Notes | EduRev

## JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2021.
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Illustration 1: There are four machines and it is known that exactly two of them are faulty. They are tested one by one in a random order till both the faulty machines are identified. Then what is the probability that only two tests are needed? (1998)
Solution: The probability that only two tests are required is the same as the probability that the first machine tested is faulty x probability that the second machine tested is faulty
= 2/4 . 1/3
= 1/6.

Illustration 2: A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is almost 0.3. Is it possible that the probability of B getting selected is 0.9?
Solution: Let P (A) and P (B) be the probabilities of selection of A and B respectively.
P (A) = 0.5, P (A∩B) ≤ 0.3
P(A∪B) = P(A) + P(B) – P(A∩B) ≤ 1
P(B) ≤ 1 + P(A∩B) – P(A)
≤ 1 + 0.3 – 0.5 ≤ 0.8
Hence, the probability of selection of B cannot be 0.9.

Illustration 3: In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it is 1/8. Find the probability that he knew the answer to the questions given that he correctly answered it.(1991)
Solution: Let us use the following symbols for denoting the various options
G for guesses
C for copies
K denotes the possibility that the examinee knows
R if the answer is right
So, P(G) = 1/3
P(C) = 1/6
P(K) = 1- (1/3 + 1/6) =1/2
Now R = (R∩G) ∪ (R∩C) ∪ (R∩K)
P(R) = P(G) P(R/G) + P(C) P(R/C) + P(K) P(R/K) … (1)
Now, P(R/G) = 1/4
P(R/C) =1/8
P(R/K) =1
Putting this in equation (1), we obtain
P(R) = 1/3. 1/4 + 1/6.1/8 + 3/6. 1
= 1/12 +1/48 + 3/6
= 29/48
Hence, the required probability = P(K∩R)/ P(R)
= 24/29

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## Mock Test Series for JEE Main & Advanced 2021

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