Four gunslingers are standing in a circle ready to shoot each other.

Question

Four gunslingers are standing in a circle ready to shoot each other. Each of them has only one bullet and will hit a chosen target with 100% probability. Each gunslinger randomly chooses one of the other gunslingers and fires. What is the probability that exactly two of them are hit ?

A. 8/27
B. 4/27
C. 3/27
D. 4/81
E. 1/81

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Attachment:

Gunslinger.jpg

chetan2u · Answer

Let A, B, C and D be the gunslingers.

(1) TWO of them, say A and B, are shot. 
C and D have a probability of choosing A or B as  \frac{2}{3} , while A and B have a probability of  \frac{1}{3}  for choosing each other.
P= \frac{2}{3} * \frac{2}{3} * \frac{1}{3} * \frac{1}{3} *1= \frac{4}{81} 
However, ways of choosing any two out of A, B, C and D is 4C2 = 6
Final P = 6* \frac{4}{81}  =  \frac{8}{27}

A

rajtare · Answer

Exactly two people will be left if two people shoots the same person, and remaining two people shoots the same person. E.g. Let 4 gunslingers be A,B,C, and D. so exactly two people will survive if A, B, Shoots C, and C, and D shoots B. Then A, and B will be alive

We can use this knowledge to build the equation. a group of two people can be formed in (4C2)/2! ways. i.e = 3
we divide here by 2! because AB =BA, CA=AC is a same pair.

Two scenarios are possible here:
1) When the each person in the group shoots the same person from the opposite group

Now each of these groups 1 person from the opposite group can be selected in 2c1 ways, and there two such group thus required probability will be
(2c1*2c1)/3^4 = 4/81

2) When 1 person shoots other person in its group, and 1 person from opposite group also shoots the same person in the group

now 1 person from the two person group can be selected in 2C1 ways, and since there are two such group thus required probability will be
(2c1*2c1)/3^4 = 4/81

thus overall probability will be 3*((4/81)+(4/81)) = 8/27

Hence correct answer should be A

demawtf · Answer

Four gunslingers are standing in a circle ready to shoot each other. Each of them has only one bullet and will hit a chosen target with 100% probability. Each gunslinger randomly chooses one of the other gunslingers and fires. What is the probability that exactly two of them are hit ?

A. 8/27
B. 4/27
C. 3/27
D. 4/81
E. 1/81

Show Spoiler

Attachment:

Gunslinger.jpg

A. 8/27
B. 4/27
C. 3/27
D. 4/81
E. 1/81

To have exactly two gunslingers hit, we need:

- i.e.; A & B to shoot each other
A-->B P:1/3
B-->A P:1/3
Combined: 1/3 * 1/3 =1/9

- C to shoot A/B and D to shoot A/B
C-->A/B P:2/3
D-->A/B P:2/3
Combined: 2/3 * 2/3 = 4/9

Total Probably of the two independent events: 1/9 * 4/9 = 4/81

Ways:
n!/(n-k)! k!

4!/2!2! = 6 ways

Since there are 6 ways to choose pairs of gunslingers to be hit, the total probability is:

6 * 4/81 = 8/27

Kinshook · Answer

Four gunslingers are standing in a circle ready to shoot each other. Each of them has only one bullet and will hit a chosen target with 100% probability. Each gunslinger randomly chooses one of the other gunslingers and fires. What is the probability that exactly two of them are hit ?

A. 8/27
B. 4/27
C. 3/27
D. 4/81
E. 1/81

What is the probability that exactly two of them are hit ?

The probability that exactly two of them are hit =  4C2 * (\frac{2}{3})^2 (\frac{1}{3})^2 = 6* \frac{4}{9} * \frac{1}{9} = \frac{8}{27}

IMO A

Four gunslingers are standing in a circle ready to shoot each other.

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Four gunslingers are standing in a circle ready to shoot each other.

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