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

BTGmoderatorDC wrote: ↑

Wed Jan 15, 2020 5:03 pm

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

Assuming that the coin is fair; thus, the probability of getting a head on a flip = 1/2 and the probability of getting a tail on a flip = 1/2

For Kate to have more than $10 but less than $15, he must win 3 or 4 times out of a total of 5 times. He must not win all 5 flips as this will make his sum = $15.

Probability of getting 3 heads out of 5 flips = 5C3*(1/2)^3*(1/2)^2 = 5C2*(1/2)^5 = (5.4/1.2)/32 = 5/16
Probability of getting 4 heads out of 5 flips = 5C4*(1/2)^4*(1/2) = 5C1*(1/2)^5 = (5)/32 = 5/32

Thus, the probability of getting 3 or 4 heads out of 5 flips = 5/16 + 5/32 = 15/32

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorDC wrote: ↑

Wed Jan 15, 2020 5:03 pm

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

In order for Kate to have more than 10 dollars but less than 15 dollars, the outcome must contain at least one head, and more tails than heads must land. Thus, either of the following two outcomes must have occurred:

T-T-T-T-H, so Kate would have 13 dollars

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 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 is 10/32.

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

Answer: D