bb wrote:

Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?

тАв 0.2

тАв 0.25

тАв 0.4

тАв 0.8

тАв 0.96

Here are 3 ways of looking at this:

(1)

WE are already given that E is chosen, so the problem is simply what is the prob that she will pick B if she chooses 2 subjects.

This is the probability that she will either choose the subject as the first pick OR the subject as the second pick.

Pr(b 1st) = 1/5

Pr(b 2nd) = Pr(b not 1st AND b 2nd) = 4/5 * 1/4 = 1/5

These 2 are mutually exclusive, hence the probability is 1/5 + 1/5 = 2/5 = 0.4

Method 2:

The probability that she will NOT leave B as one of her 3 unpicked choice is exactly the same problem.

Pr(B is NOT one of 3 unpicked choices) = 4/5 * 3/4 * 3/2 = 2/5 = 0.4

Method 3:

Pr (b picked in 2 choices) = 1 - Pr(B not picked of her two choices)

= 1 - (4/5 * 3/4) = 1 - 3/5 = 2/5 = 0.4.

_________________

Best,

AkamaiBrah

Former Senior Instructor, Manhattan GMAT and VeritasPrep

Vice President, Midtown NYC Investment Bank, Structured Finance IT

MFE, Haas School of Business, UC Berkeley, Class of 2005

MBA, Anderson School of Management, UCLA, Class of 1993