snorkeler wrote:

Josh and Dan have a 12 apples each. Together they flip a coin 6 times. For every heads, Josh receives an apple from Dan, and for every tails Dan receives an apple from Josh. After the coin has been flipped 6 times, what is the probability that Josh has more than 12 apples but fewer than 18?

(A) 1/64

(B) 15/64

(C) 21/64

(D) 21/32

(E) 5/6

Let us start with counting the number of apples Josh can have that satisfies the condition that "Josh has more than 12 apples but fewer than 18"

13 - Not possible. Since he gets one apple for every heads and gives one away for every heads. There is no scenario where his "net gain" would be +1 apple if the coin is only being tossed 6 times.

14 - Possible for 4 Heads and 2 Tails OR for 4 Tails and 2 Heads

15 - Not possible. Same as 13.

16 - Possible for 5 Heads and 1 Tail OR 5 Tails and 1 Head

17 - Not possible.

So the 4 cases where this scenario is possible are - [4H2T + 4T2H + 5H1T + 5T1H]

No. of ways to get 4H and 2T in 6 tosses = HHHHTT, HHTTHH, and so on, so to calculate this - 6!/2!4! = 15 and prob. of getting either head or tails = 1/2, so 15*(1/2) = 15/2

No. of ways to get 4T and 2H = 15 x Prob of H or T = 15*(1/2) = 15/2

No. of ways to get 5H and 1T = 6!/5! x 1/2

No. of ways to get 5T and 1 H x Prob of getting head or tails = 6!/5! x 1/2

Adding all 4 we get 21

Total possibilities of H or T with 6 tosses = 2x2x2x2x2x2 = 64

Therefore, 21/64