Hi All,

This is a layered probability question that requires some additional math (the combination formula, or the equivalent "mapping" of all possibilities).

We're told that the coin will be flipped 5 times; with every "win", Kate gets $1 and with every "loss", Kate loses $1. Kate starts with $10. We're asked for the probability that Kate ends up with MORE than $10 but less than $15 after 5 tosses.

Let's start with the total number of possible outcomes. Since each coin has 2 options (heads or tails), there are 2^5 = 32 possible outcomes for the 5 flips (which will include a certain number of similar outcomes in different orders - for example HHTTT and THTHT).

To end up with MORE than $10, Kate has to win MORE tosses than she loses. However, if she were to win all 5 tosses, she'd have $15 and we want her to end up with LESS than $15. This means that Kate has to win EITHER 3 times or 4 times.

Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:

Combinations = N!/[K!(N-K)!]

For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins

For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins

So, there are a total of 10+5 = 15 combinations of 5 tosses that "fit" what we're looking for and 32 possible outcomes total.

Final Answer:

GMAT assassins aren't born, they're made,

Rich

I'm confused about something.

there are 2^5 = 32 possible......this means the order is important (matter) so we deal with permutation in nature.

But you used combinations for 5C3 & 5C4.

How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator).

My questions to you stems from a reply to me from another tutor/expert in another question as below:

I hope I clarified my point and where the confusion happened.