Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives

[Gmat_mission]

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

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT

[deloitte247]

For Kate to have more than $10 and Danny to have less than $15 the fair coin must land on head 3 or 4 times
However, the fair coin has 2 sides and it is flipped 5 times
$$Total\ number\ of\ possibilities\ =\ 2^5=32$$
$$probability\ of\ getting\ 3\ heads\ =\ \frac{5C3}{32}$$
$$where\ 5C3=\frac{5!}{3!\left(5-3\right)!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\left(2\cdot1\right)}$$
$$=\frac{5\cdot4}{2\cdot1}=\frac{20}{2}=10$$
$$\Pr obability\ of\ getting\ 3\ heads\ =\frac{10}{32}$$
$$\Pr obability\ of\ getting\ 4\ heads\ =\frac{5C4}{3}$$
$$where\ 5C4=\frac{5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1\left(5-4\right)!}$$
$$=\frac{5}{1!}=5$$
$$probability\ of\ getting\ 4\ heads=\frac{5}{32}$$
$$probability\ of\ getting\ 3\ or\ 4\ heads=\frac{10}{32}+\frac{5}{32}$$
$$=\frac{10+5}{32}$$
$$=\frac{15}{32}$$
$$Answer\ =D$$

[Scott@TargetTestPrep]

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

[spoiler]OA=D[/spoiler]

Solution:

In order for Kate to have more than 10 dollars but less than 15 dollars, either of the following two outcomes must have occurred:

First outcome: T-T-T-T-H, so Kate would have 13 dollars

Second outcome: T-T-T-H-H, so Kate would have 11 dollars

Let’s calculate the probability of each outcome:

P(T-T-T-T-H) = (1/2)^5 = 1/32

Since T-T-T-T-H can be arranged in 5!/4! = 5 ways, the probability of the first outcome is 5/32.

Next:

P(T-T-T-H-H) = (1/2)^5 = 1/32

Since T-T-T-H-H can be arranged in 5!/(3! x 2!) = 10 ways, the probability of the second outcome is 10/32.

So the overall probability that Kate has more than $10 but less than $15 is 15/32.

Answer: D