7) Suppose that you roll a pair of standard 6-sided dice a total of 42 times. What is the probability that you will get a sum of 5 at least four times?

Answer 1

The probability of getting 5 on one toss of a pair is 1/9. The most probable number of fives you will get after 42 tosses is 4 or 5.

Add the probabilities of getting 5 four, five, six, seven, eight etc timesout of 42 tosses. By the time you get to 15 times, the probability will be very small and you can quit adding them up.

The probability of rolling 5 four times is (1/9)^4*(8/9)^38*42!/(38!*4!)
= 119,930*1.524*10^-4*1.138*10^-2
= 0.20803

The probability of rolling 5 five times is (1/9)^5*(8/9)^37*42!/(37!*5!)
= 8.507*10^5*1.6935*10^-5*1.2804*10^-2
= 0.18447

I am not going to continue this because it would take half the night, but you get the idea.

There is an easier way to do this that just occurred to me. It will give the same answer. Here it is:
Add the probabilities of getting 5 zero, one, two or three times, and subtract that from 1.

The probability of never getting five is (8/9)^42 = 0.007105
The probability of getting 5 only once is (1/9)*(8/9)^41*42 = 0.037304
The probability of getting 5 twice is
(1/9)^2*(8/9)^40*(42*41*/2) = 0.095591
Now YOU compute the probability of getting 5 three times. Add that to the probabilites above and subtract the sum from one. You should end up with a number around 0.6.

That will be the answer.