Kate and Danny each have $10. Together, they flip a fair coin 5 times. : Problem Solving (PS)

powerka

Joined: 22 Jul 2009

Last visit: 29 Apr 2016

Posts: 100

Own Kudos:

549

[3]

Given Kudos: 18

Posts: 100

Kudos: 549

[3]

22 Sep 2009, 14:40

3

Kudos

Bookmarks

Decided to post this question as Manhattan's explanation was unnecessarily cumbersome.

Question asks for the probability of the coin landing tails up either 3 or 4 times = P(3t) + P(4t)

Binomial distribution formula: nCk p^k (1-p)^(n-k)

P(3t) = 5C3 (1/2)^3 (1/2)^2 = 10 (1/2)^5
P(4t) = 5C4 (1/2)^4 (1/2)^1 = 5 (1/2)^5

=> P(3t) + P(4t) = 15/32

deep5586

Joined: 18 Oct 2011

Last visit: 28 Nov 2011

Posts: 2

Own Kudos:

1

[1]

Concentration: Entrepreneurship, Marketing

GMAT Date: 11-25-2011

GPA: 3.9

Posts: 2

Kudos: 1

[1]

19 Oct 2011, 09:26

1

Kudos

Bookmarks

y arent we taking into account the probablility of 2 heads or 1

heman2402

Joined: 12 Jan 2011

Last visit: 30 May 2014

Posts: 35

Own Kudos:

74

[14]

Given Kudos: 5

Status:Next engagement on Nov-19-2011

Location: New Delhi, India

Schools:IIM, ISB, & XLRI

Q45 V21 GMAT 2: 630 Q46 V31

WE 1: B.Tech (Information Technology)

Posts: 35

Kudos: 74

[14]

19 Oct 2011, 11:20

13

Kudos

1

Bookmarks

deep5586

y arent we taking into account the probablility of 2 heads or 1

For 2 & 1 heads Kate will end up with < $10 and we want her to win :-)

. Therefore, only possibilities are 3 or 4 heads.

I made an educated guess and it worked fine. :-D

Ans- 'D'

MGMAT's anagram helped here as well.

HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10

Total acceptable cases = 15
Total cases = 32

P = 15/32 :-D

manpreetsingh86

Joined: 13 Jun 2013

Last visit: 19 Dec 2022

Posts: 219

Own Kudos:

1,147

[14]

Given Kudos: 14

Posts: 219

Kudos: 1,147

[14]

31 Oct 2014, 02:56

11

Kudos

1

Bookmarks

Bunuel

Tough and Tricky questions: Probability.

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

total number of possible outcomes when coin will be flipped 5 times= 2^5=32

now, for kate to have more than $10, she must have min. 3 favorable and max. 4 favorable draws in the 5 throws. (because if she wins in all the 5 draws, then she will have $15)

kate will win, every time tails (T) appears on the coin.

case 1) kate wins in 3 draws and losses in two. so we have 3 T's and 2H's(T,T,T,H,H), which can arrange themselves in 5!/(3!)(2!)=10

case 2) kate wins in 4 draws and losses in the 1 draw. so we have 4 T's and 1H (T,T,T,T,H), which can arrange themselves in 5!/4!=5

thus total no. of favorable ways=10+5=15
and thus required probability=15/32

nitinagg29

Joined: 07 May 2015

Last visit: 29 Dec 2015

Posts: 8

Own Kudos:

14

[1]

Given Kudos: 9

Location: India

Concentration: International Business, Finance

Schools: Melbourne '19 NUS '18 Kellogg '18 Ross '18 Thunderbird '16 (A)

GPA: 3.5

WE:Information Technology (Computer Software)

Schools: Melbourne '19 NUS '18 Kellogg '18 Ross '18 Thunderbird '16 (A)

Posts: 8

Kudos: 14

[1]

29 Jun 2015, 07:45

1

Kudos

Bookmarks

Solved it using Pascal's triangle method and able to got the answer in 30 sec. https://en.wikipedia.org/wiki/Pascal's_triangle

Hector1S

Joined: 31 May 2017

Last visit: 05 Apr 2023

Posts: 30

Own Kudos:

12

[1]

Given Kudos: 67

Products:

Posts: 30

Kudos: 12

[1]

22 Jun 2018, 23:49

1

Kudos

Bookmarks

Can anybody explain why the total no.of possibilities is 32 and not 10.

Ramyahari

Joined: 18 Apr 2018

Last visit: 03 Sep 2018

Posts: 1

Own Kudos:

2

[2]

Given Kudos: 6

Posts: 1

Kudos: 2

[2]

30 Jul 2018, 22:15

2

Kudos

Bookmarks

My explanation is
Total outcomes of tossing a coin =2^5 =32

So the probability is exactly half times head and half time tails...i.e.16/32
And since it is asked for 10-15
Answer is 15/32.

Is my approach right??

Posted from my mobile device

ScottTargetTestPrep

Target Test Prep Representative

Joined: 14 Oct 2015

Last visit: 15 Jul 2025

Posts: 21,112

Own Kudos:

26,166

[4]

Given Kudos: 296

Status:Founder & CEO

Affiliations: Target Test Prep

Location: United States (CA)

Expert

Expert reply

Posts: 21,112

Kudos: 26,166

[4]

06 Feb 2019, 19:32

2

Kudos

2

Bookmarks

powerka

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

In order for Kate to have more than 10 dollars but less than 15 dollars, 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

exc4libur

Joined: 24 Nov 2016

Last visit: 22 Mar 2022

Posts: 1,702

Own Kudos:

1,431

[1]

Given Kudos: 607

Location: United States

Posts: 1,702

Kudos: 1,431

[1]

23 Aug 2019, 06:10

Kudos

1

Bookmarks

powerka

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

cases in which $10<total<$15:
WWWLL=+3-2=+1… total=$11… prob: $3W2L=(1/2)^5•{arrangements}\frac{5!}{3!2!}=\frac{10}{32}$
WWWWL=+4-1=+3… total=$13… prob: $4W1L=(1/2)^5•{arrangements}\frac{5!}{4!1!}=\frac{5}{32}$
sum cases: $\frac{10}{32}+\frac{5}{32}=\frac{15}{32}$

Answer (D)

kunalc20

Joined: 17 May 2016

Last visit: 10 Apr 2022

Posts: 20

Own Kudos:

9

Given Kudos: 33

GMAT 1: 590 Q47 V25 Verified score

Posts: 20

Kudos: 9

06 Jun 2020, 12:54

Kudos

Bookmarks

manpreetsingh86

Bunuel

Tough and Tricky questions: Probability.

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

total number of possible outcomes when coin will be flipped 5 times= 2^5=32

now, for kate to have more than $10, she must have min. 3 favorable and max. 4 favorable draws in the 5 throws. (because if she wins in all the 5 draws, then she will have $15)

kate will win, every time tails (T) appears on the coin.

case 1) kate wins in 3 draws and losses in two. so we have 3 T's and 2H's(T,T,T,H,H), which can arrange themselves in 5!/(3!)(2!)=10

case 2) kate wins in 4 draws and losses in the 1 draw. so we have 4 T's and 1H (T,T,T,T,H), which can arrange themselves in 5!/4!=5

thus total no. of favorable ways=10+5=15
and thus required probability=15/32

Why do we need to consider the arrangement here
Please help. its all about getting heads and tails . We need to have 4 heads or 3 heads for kate to have money more than 10 and less than 15
doesnt matter at which turn we are getting head or tail

adityaprateek15

Joined: 26 May 2023

Last visit: 15 July 2025

Posts: 127

Own Kudos:

80

Given Kudos: 232

Location: India

Schools: Ross '26 Tepper '26 McCombs '26

GPA: 2.7

Schools: Ross '26 Tepper '26 McCombs '26

Posts: 127

Kudos: 80

23 May 2025, 12:54

Kudos

Bookmarks

To solve this question, the key is understanding the impact of each coin flip:

If Kate wins once, she also loses four times (since there are 5 total flips), which results in:
10+1−4=7

So, each win adds +1 to her amount, and each loss subtracts 1.
We’re told that Kate starts with $10 and we want the total after 5 flips to be more than 10 but less than 15.
This means her net gain must be between 1 and 4, i.e., she must win more times than she loses.

So, the only conditions are:

1.Kate wins 3 times and looses 2 other times:5c3 x (1/2)^3 x (1/2)^2 = 10/32

2.Kate wins 4 times and loose 1 other time: 5c4 x (1/2)^4 x (1/2)^1 = 5/32

Total = 10+5/32 = 15/32

powerka

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

Source: Manhattan GMAT Archive (tough problems set).doc

Kate and Danny each have $10. Together, they flip a fair coin 5 times.

Prep Toolkit

My Rewards

GMAT Problem Solving (PS) Questions

Kate and Danny each have $10. Together, they flip a fair coin 5 times.

Prep Toolkit

My Rewards