Ok, I got two problems that I am stuck on.

1. If the letters in the word POKER are rearranged, what is the probability that the word will begin with the letter E and end with the letter O?
Answer: _________


2. A box contains 80 balls numbered from 1 to 80. If 13 balls are drawn with replacement, what is the probability that at least two of them have the same number?
Answer: ______________


I dont need the answer, because I need to learn this for the test, I really just need to know how to set up the problem. On the first problem, I thought I could set it up like ( C(5,1)+C(4,1)+C(3,3) ) / C(5,5)
My friend told me that, but I can't get the right answer with that.

If someone can tell me just how to set up the problem it will be greatly appreciated! Thanks guys!!

 

On POKER, seems like there is a 20% chance that on a random rearrange that the first letter will be an "e" (1 in 5) and a 25% chance that of the four remaining letters, the last one will be an O (1 in 4) , so .20 x .25 = .05, so 5% chance of ending up EXXXO.

 

Thanks! I guess I didn't need all those Combination formulas.

 

Another way to look at the POKER problem, using your combination (actually, for this problem, permutation) formula is this:

In general, the probability is the number of successful outcomes divided by the number of possible outcomes.

For the number of possible arrangements, you are permuting five letters taken five at a time: P(5,5).

For the number of successful arrangements you have fixed two letters (E and O), so you are only permuting the remaining three letters taken three at a time: P(3,3).

Thus:

prob = P(3,3) / P(5,5) = 6 / 120 = 0.05.

For the ball problem, the number of possible outcomes is 80^13.

The number of successful outcomes is very difficult to compute directly. However, it is very easy to compute the number of unsuccessful outcomes: the number of ways in which no two balls match. To get no matches you have 80 possible choices for the first ball, 79 for the second, 78 for the third, and so on down to 68 for the thirteenth. Thus, the number of ways to fail is 80 * 79 * . . . * 68 = 80! / 67!. Therefore, the number of ways to succeed is 80^13 - 80! / 67!.

prob = (80^13 - 80! / 67!) / (80^13) = 0.643.

 

Magician, you took the words right out of my mouth!

 

Thanks!!

 

Be sure that what you're understanding is the thinking (all of it) that led to how to count the successful outcomes and possible outcomes, rather than merely trying to remember the formula for the answer. The next problem may require a different approach to the counting, thus a substantially different formula.

 

I find this much more satisfying than plagiarizing them later.