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 fivecoin 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
Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives
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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
$$Total\ number\ of\ possibilities\ =\ 2^5=32$$
$$probability\ of\ getting\ 3\ heads\ =\ \frac{5C3}{32}$$
$$where\ 5C3=\frac{5!}{3!\left(53\right)!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\left(2\cdot1\right)}$$
$$=\frac{5\cdot4}{2\cdot1}=\frac{20}{2}=10$$
$$\Pr obability\ of\ getting\ 3\ heads\ =\frac{10}{32}$$
$$\Pr obability\ of\ getting\ 4\ heads\ =\frac{5C4}{3}$$
$$where\ 5C4=\frac{5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1\left(54\right)!}$$
$$=\frac{5}{1!}=5$$
$$probability\ of\ getting\ 4\ heads=\frac{5}{32}$$
$$probability\ of\ getting\ 3\ or\ 4\ heads=\frac{10}{32}+\frac{5}{32}$$
$$=\frac{10+5}{32}$$
$$=\frac{15}{32}$$
$$Answer\ =D$$
However, the fair coin has 2 sides and it is flipped 5 times
$$Total\ number\ of\ possibilities\ =\ 2^5=32$$
$$probability\ of\ getting\ 3\ heads\ =\ \frac{5C3}{32}$$
$$where\ 5C3=\frac{5!}{3!\left(53\right)!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\left(2\cdot1\right)}$$
$$=\frac{5\cdot4}{2\cdot1}=\frac{20}{2}=10$$
$$\Pr obability\ of\ getting\ 3\ heads\ =\frac{10}{32}$$
$$\Pr obability\ of\ getting\ 4\ heads\ =\frac{5C4}{3}$$
$$where\ 5C4=\frac{5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1\left(54\right)!}$$
$$=\frac{5}{1!}=5$$
$$probability\ of\ getting\ 4\ heads=\frac{5}{32}$$
$$probability\ of\ getting\ 3\ or\ 4\ heads=\frac{10}{32}+\frac{5}{32}$$
$$=\frac{10+5}{32}$$
$$=\frac{15}{32}$$
$$Answer\ =D$$
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Solution:Gmat_mission wrote: ↑Wed Jun 24, 2020 8:01 amKate 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 fivecoin 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]
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: TTTTH, so Kate would have 13 dollars
Second outcome: TTTHH, so Kate would have 11 dollars
Let’s calculate the probability of each outcome:
P(TTTTH) = (1/2)^5 = 1/32
Since TTTTH can be arranged in 5!/4! = 5 ways, the probability of the first outcome is 5/32.
Next:
P(TTTHH) = (1/2)^5 = 1/32
Since TTTHH 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
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