Last visit was: 01 May 2026, 11:43 It is currently 01 May 2026, 11:43
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,521
 [14]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,521
 [14]
20 men are held captive by a pirate lord, including two friends Jack a 06 May 2019, 15:02
Kudos
Add Kudos
13
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

21% (02:55) correct 79% (02:24) wrong based on 47 sessions

History

Date
Time
Result
Not Attempted Yet
20 men are held captive by a pirate lord, including two friends Jack and Tony. The pirate decides to set 10 of them free. The men are randomly divided into 10 pairs. Each pair of men then flip a fair coin to decide who goes free.The probability that both Jack and Tony are set free is A/B where A and B are co-prime positive integers. Find the value of A+B?

A. 5
B. 15
C. 38
D. 45
E. 47
ShowHide Answer
Official Answer
E
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,521
 [2]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,521
 [2]
20 men are held captive by a pirate lord, including two friends Jack a 07 May 2019, 11:15
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If you want both Jack and Tony to be free, they must be in different pair and both must win.
The probability of Jack and Tony to be in different pair- 18/19 (Tony or Jack could be pair with 19 others)
The probability of both will win in their respective pair- (1/2)*(1/2) (It does not matter who else will win in other pairs)
The probability that both Jack and Tony are set free- (18/19)*(1/4)=9/38
A+B = 9+38= 47
avatar
kukretipiyush
avatar
Joined: 08 Oct 2018
Last visit: 16 Jul 2019
Posts: 8
Own Kudos:
4
Given Kudos: 9
Posts: 8
Kudos: 4
20 men are held captive by a pirate lord, including two friends Jack a 09 May 2019, 21:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nick1816
If you want both Jack and Tony to be free, they must be in different pair and both must win.
The probability of Jack and Tony to be in different pair- 18/19 (Tony or Jack could be pair with 19 others)
The probability of both will win in their respective pair- (1/2)*(1/2) (It does not matter who else will win in other pairs)
The probability that both Jack and Tony are set free- (18/19)*(1/4)=9/38
A+B = 9+38= 47
Show more

Can you explain how did you get probability of Jack and Tony to be in different pair as 18/19.
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,521
 [2]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,521
 [2]
20 men are held captive by a pirate lord, including two friends Jack a 09 May 2019, 21:59
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
When there are no restrictions. Suppose Jack choose his partner first. Then he has 19 options. Now 18 pirates left, now any random guy among them has 17 options to choose his partner...and this goes on till last 2 pirates left and any of the last 2 pirates has only one option left.
Total number of ways to divide them in 10 pair= 19*17*15*13......*5*3*1
When Jack and Tony has to be in different pair. Now Jack has 18 options to select his partner, as he can't select Tony. The next Pirate has same 17 options. Rest process will remain same .
Total number of ways to divide them in 10 pair, when Jack and Tony has to be in different pair= 18*17*15.....*5*3*1

The probability of Jack and Tony to be in different pair-(18*17*15.....*5*3*1)/(19*17*15*13......*5*3*1)=18/19


kukretipiyush
nick1816
If you want both Jack and Tony to be free, they must be in different pair and both must win.
The probability of Jack and Tony to be in different pair- 18/19 (Tony or Jack could be pair with 19 others)
The probability of both will win in their respective pair- (1/2)*(1/2) (It does not matter who else will win in other pairs)
The probability that both Jack and Tony are set free- (18/19)*(1/4)=9/38
A+B = 9+38= 47

Can you explain how did you get probability of Jack and Tony to be in different pair as 18/19.
Show more
avatar
ailyninja
avatar
Joined: 06 May 2019
Last visit: 28 Feb 2020
Posts: 6
Own Kudos:
4
Given Kudos: 18
Location: China
Schools: Stanford '22 Anderson '22 Booth '22 Sloan '22 Yale '22
GPA: 3.77
Schools: Stanford '22 Anderson '22 Booth '22 Sloan '22 Yale '22
Posts: 6
Kudos: 4
20 men are held captive by a pirate lord, including two friends Jack a 24 May 2019, 07:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kukretipiyush
nick1816
If you want both Jack and Tony to be free, they must be in different pair and both must win.
The probability of Jack and Tony to be in different pair- 18/19 (Tony or Jack could be pair with 19 others)
The probability of both will win in their respective pair- (1/2)*(1/2) (It does not matter who else will win in other pairs)
The probability that both Jack and Tony are set free- (18/19)*(1/4)=9/38
A+B = 9+38= 47

Can you explain how did you get probability of Jack and Tony to be in different pair as 18/19.
Show more

Besides Jack, there are 19 men. 18 out of 19 men (except for Tony) can be paired with Jack --> 18/19. We don't need to consider how the rest 18 of them are paired, because we're sure that Jack and Tony are not a pair by now.
User avatar
hadimadi
User avatar
Joined: 26 Oct 2021
Last visit: 03 Dec 2022
Posts: 113
Own Kudos:
31
Given Kudos: 94
Posts: 113
Kudos: 31
20 men are held captive by a pirate lord, including two friends Jack a 22 Dec 2021, 04:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

I solved it as follows:

There are 20C2=190 unique possibilities to create pairs of two. There are (190)*(152)*(114)*(76)*(38) total possibilities for different groups created out of these pairs (where order does matter, for now). Why? We start to pick 1 out of 190, and if we pick for example (a,b), then all other two pairs which contain EITHER a OR b have to go out. There are in total 19 pairs with a, 19 pairs with b. Meaning we have 38 less to choose from, and 190-38=152. This process continues until we have done it 4 times and reach 38.
We can determine the number of possibilities in which Jack and Tony ARE in a group by 5C2*[(152)*(114)*(76)*(38)]. Why 5C2? Because as order DOES matter for out possibilities we calculated above, it could be that the pair (Jack,Tony) is the first one, second one, first and third one,..., and so on! In total we get
Possibilities where both in a group/total=[5C2*152*...*38)]/[(190*152*...*38)]=5C2/190=10/190.
It follows that in 1-(10/190)=18/19 cases they AREN'T in a group. The probability of both to win in their respective groups is 1/4, and in total we get 18/19*(1/4)=9/38=A/B, and with A/B being irreducible (co-prime)-> A+B=9+38=47.
Moderators:
Bunuel
Math Expert
109999 posts
Krunaal
Tuck School Moderator
852 posts