• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives

##### This topic has expert replies
Legendary Member
Posts: 563
Joined: 01 Mar 2018
Followed by:1 members

### Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives by Gmat_mission » Wed Jun 24, 2020 8:01 am ## Timer 00:00 ## Your Answer A B C D E ## Global Stats 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

Legendary Member
Posts: 1846
Joined: 02 Mar 2018
Followed by:3 members

### Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate give by deloitte247 » Sat Jun 27, 2020 2:26 am 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 $$Total\ number\ of\ possibilities\ =\ 2^5=32$$ $$probability\ of\ getting\ 3\ heads\ =\ \frac{5C3}{32}$$ $$where\ 5C3=\frac{5!}{3!\left(5-3\right)!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\left(2\cdot1\right)}$$ $$=\frac{5\cdot4}{2\cdot1}=\frac{20}{2}=10$$ $$\Pr obability\ of\ getting\ 3\ heads\ =\frac{10}{32}$$ $$\Pr obability\ of\ getting\ 4\ heads\ =\frac{5C4}{3}$$ $$where\ 5C4=\frac{5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1\left(5-4\right)!}$$ $$=\frac{5}{1!}=5$$ $$probability\ of\ getting\ 4\ heads=\frac{5}{32}$$ $$probability\ of\ getting\ 3\ or\ 4\ heads=\frac{10}{32}+\frac{5}{32}$$ $$=\frac{10+5}{32}$$ $$=\frac{15}{32}$$ $$Answer\ =D$$ ### GMAT/MBA Expert GMAT Instructor Posts: 5170 Joined: 25 Apr 2015 Location: Los Angeles, CA Thanked: 43 times Followed by:23 members ### Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate give

by Scott@TargetTestPrep » Mon Jun 29, 2020 5:56 am
Gmat_mission wrote:
Wed Jun 24, 2020 8:01 am
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.