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nick1816
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20 men are held captive by a pirate lord, including two friends Jack a 06 May 2019, 15:02
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21% (02:55) correct 79% (02:24) wrong based on 47 sessions

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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
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nick1816
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20 men are held captive by a pirate lord, including two friends Jack a 07 May 2019, 11:15
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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
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20 men are held captive by a pirate lord, including two friends Jack a 09 May 2019, 21:28
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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
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Can you explain how did you get probability of Jack and Tony to be in different pair as 18/19.
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20 men are held captive by a pirate lord, including two friends Jack a 09 May 2019, 21:59
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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.
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20 men are held captive by a pirate lord, including two friends Jack a 24 May 2019, 07:33
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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.
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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.
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20 men are held captive by a pirate lord, including two friends Jack a 22 Dec 2021, 04:26
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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.
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