A canditate takes three tests in succession and the probability of passing the first test is p . the probability of passing each succeeding test is p or p/2 according as he passes or fails in the preceeding one. the candidate is selected if he passes atleast 2 tests. the probability that the candidate is selected is ?

2

Answers

The Brainliest Answer!
2015-04-21T21:36:56+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Notice that if p is the probability of pass, then the probability of fail is 1-p.

The candidate can pass atleast 2 tests in 4 ways and the probabilities are given by :
1) passing all 3 tests : p*p*p=p^3
2) passing first two tests and failing third test : p*p*(1-p)=p^2(1-p)
3) passing first test, failing second and passing third : p*(1-p)*p/2=p^2(1-p)/2
4) failing first test, passing second and third : (1-p)*p/2*p=p^2(1-p)/2

Add them up to get the probability for candidate to be selected :
p^3+p^2(1-p)+p^2(1-p)/2+p^2(1-p)/2=2p^2-p^3
2 5 2
  • Brainly User
2015-04-21T21:38:32+05:30
We need to know the probability that he/she passes at least two tests. 
Let the three tests be named as A, B and C (in order). The required probability is 
 P(A \cap B \cap C)+P(A' \cap B \cap C)+P(A \cap B' \cap C)+P(A \cap B \cap C') \\ =p.p.p + (1-p).\frac{p}{2}.p +p.(1-p).\frac{p}{2} + p.p.(1-p) \\ = p^{3}+2p^{2}-2p^{3} \\ =2p^{2}-p^{3} where we have used P(R \cap S)=P(R).P(S|R).
I am not so sure about it, in any case feel free to report it.

1 5 1