Probability - Random Variables

[AsianMan]

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or an "off" day. He figures that if he has an on day, then each of his examiners will pass him independently of each other, with probability .8, whereas if he has an off day, this probability will be reduced to .4. Suppose that the student will pass the examinatorion if a majority of the examiners pass him. If the student feels that he is twice as likely to have an off day as he is to have an on day, should he request an examination wiht 3 examiners or 5?

_________
I thought this was a binomial probability distriubtion question, with possibility of success = .4*(2/3) + .8*(1/3) = 8/15

I totaled the probabilities for 3,4, and 5 successes for a 5 examiner group and 2 and 3 successes for a 3 man group and found the probability of success for the 5 man group was higher (.562315 vs .549926), but the book says you have a better chance of passing with a 3 man group.

This is the 2nd question in a row of this type I have screwed up and am desperate to see what I'm doing wrong.

 

[Billy Bob]

You did the 2/3 vs 1/3 weighting at the beginning instead of the end.

Find prob of winning with n=5 and p=0.8.
Find prob of winning with n=5 and p=0.4.
Take weighted average of those two probabilities.

Repeat using n=3 instead of n=5.