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Four robbers went to a safe place after a robbery. At midnight, each robber randomly chose one of the other three robbers and shot him. What is the probability that exactly two robbers were shot?
- a)1/2
- b)1/3
- c)1/4
- d)8/27
Correct answer is option 'D'. Can you explain this answer?
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Four robbers went to a safe place after a robbery. At midnight, each r...
To solve this problem, we can use the concept of permutations and combinations. Let's break down the problem step by step:
Step 1: Total possible outcomes
Each robber has 3 choices of who to shoot (out of the other 3 robbers). So the total number of possible outcomes is 3 * 3 * 3 * 3 = 81.
Step 2: Counting favorable outcomes
To determine the probability of exactly two robbers being shot, we need to count the number of favorable outcomes.
Case 1: Two robbers shoot each other
In this case, we need to choose any two robbers out of the four. There are 4C2 ways to do this, which is equal to 6. Once we have chosen the two robbers, there is only one way for each of them to shoot the other. Therefore, the number of favorable outcomes in this case is 6 * 1 * 1 = 6.
Case 2: One robber shoots two others
In this case, we need to choose one robber out of the four. There are 4 ways to do this. Once we have chosen the robber, there are 3C2 ways to choose the two robbers that will be shot. For each of these choices, there is only one way for the robber to shoot the two others. Therefore, the number of favorable outcomes in this case is 4 * 3C2 * 1 * 1 = 4 * 3 = 12.
Step 3: Calculate the probability
The probability is given by the number of favorable outcomes divided by the total number of possible outcomes.
Probability = (number of favorable outcomes) / (total number of possible outcomes)
Probability = (6 + 12) / 81
Probability = 18 / 81
Probability = 2 / 9
So the probability that exactly two robbers were shot is 2/9, which is equivalent to 8/27.
Therefore, the correct answer is option D) 8/27.
Step 1: Total possible outcomes
Each robber has 3 choices of who to shoot (out of the other 3 robbers). So the total number of possible outcomes is 3 * 3 * 3 * 3 = 81.
Step 2: Counting favorable outcomes
To determine the probability of exactly two robbers being shot, we need to count the number of favorable outcomes.
Case 1: Two robbers shoot each other
In this case, we need to choose any two robbers out of the four. There are 4C2 ways to do this, which is equal to 6. Once we have chosen the two robbers, there is only one way for each of them to shoot the other. Therefore, the number of favorable outcomes in this case is 6 * 1 * 1 = 6.
Case 2: One robber shoots two others
In this case, we need to choose one robber out of the four. There are 4 ways to do this. Once we have chosen the robber, there are 3C2 ways to choose the two robbers that will be shot. For each of these choices, there is only one way for the robber to shoot the two others. Therefore, the number of favorable outcomes in this case is 4 * 3C2 * 1 * 1 = 4 * 3 = 12.
Step 3: Calculate the probability
The probability is given by the number of favorable outcomes divided by the total number of possible outcomes.
Probability = (number of favorable outcomes) / (total number of possible outcomes)
Probability = (6 + 12) / 81
Probability = 18 / 81
Probability = 2 / 9
So the probability that exactly two robbers were shot is 2/9, which is equivalent to 8/27.
Therefore, the correct answer is option D) 8/27.
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Four robbers went to a safe place after a robbery. At midnight, each robber randomly chose one of the other three robbers and shot him. What is the probability that exactly two robbers were shot?a)1/2b)1/3c)1/4d)8/27Correct answer is option 'D'. Can you explain this answer?
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