The company has two positions open. They will interview candidates until two qualified candidates are found. Let Ybe the number of candidates interviewed up to and including the second qualified candidate.

Question

a. What is the smallest possible value for Y?
b. What is the probability that Y takes on that value?
c. Let X represent the number of applicants who are interviewed up to and including the first qualified candidate. Find P(Y=4 | X =1).
d. Find P(Y=4 | X=2).
e. Find P(Y=4).

Answer 1

Let Q denote a qualified candidate, and U an unqualified one.



(a) If the first two candidates are both qualified, then Y = 2. This is the smallest possible value.



(b) P(Y = 2) = P(QQ) = (0.3)2 = 0.09



(c)

Now P(Y = 4 and X = 1) = P(QUUQ) = (0.3)(0.7)(0.7)(0.3) = 0.0441, and P(X = 1) = P(Q) = 0.3.

Therefore P(Y = 4|X = 1) = 0.0441?0.3 = 0.147.



(d)

Now P(Y = 4 and X = 2) = P(UQUQ) = (0.7)(0.3)(0.7)(0.3) = 0.0441, andP(X = 2) = P(UQ) = (0.7)(0.3) = 0.21.

Therefore P(Y = 3|X = 2) = 0.0441?0.21 = 0.21.



(e) P(Y = 4)=P(QUUQ) + P(UQUQ) + P(UUQQ)

=(0.3)(0.7)(0.7)(0.3) + (0.7)(0.3)(0.7)(0.3) + (0.7)(0.7)(0.3)(0.3)

= 0.0441 + 0; 0441 + 0.0441

= 0.1323