Problem Solving

BTGmoderatorDC wrote:Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than $10 but less than $15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

For every tails, Kate gets $1; for every heads, she loses $1.

0 tails, 5 heads: Kate has 10+0-5 = $5
1 tails, 4 heads: Kate has 10+1-4 = $7
2 tails, 3 heads: Kate has 10+2-3 = $9
3 tails, 2 heads: Kate has 10+3-2 = $11
4 tails, 1 heads: Kate has 10+4-1 = $13
5 tails, 0 heads: Kate has 10+5-0 = $15

Only 3 tails and 4 tails will give Kate between $10 and $15.

Question stem, rephrased:
What is the probability that coin flipped 5 times will yield exactly 3 tails or exactly 4 tails?

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2.
P(5 tails) = 1/32.
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

BTGmoderatorDC wrote:Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than $10 but less than $15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

OA D

Source: Manhattan Prep

Since the payoffs are equal, one can quickly see that Kate has a 50% chance of having more than what she started with, $10.

But that includes the excluded case of having $15. To win an additional $5, she'd have to win every toss, which has a probability of (1/2)^5 or 1/32.

So the probability is 1/2 = 16/32 - 1/32 = [spoiler]D, 15/32[/spoiler]