From a box 𝐴A containing 3 white and 6 black balls, 5 balls are transferred into an empty box 𝐵B. Let 𝑋X be a random variable that represents the number of white balls which are transferred from 𝐴A to 𝐵B. What value of random variable will have the least probability?
Question
From a box 𝐴A containing 3 white and 6 black balls, 5 balls are transferred into an empty box 𝐵B. Let 𝑋X be a random variable that represents the number of white balls which are transferred from 𝐴A to 𝐵B. What value of random variable will have the least probability?
Solution
The random variable X represents the number of white balls transferred from box A to box B. The least number of white balls that can be transferred is 0 and the maximum is 3 (since there are only 3 white balls in box A).
Therefore, the possible values of the random variable X are 0, 1, 2, and 3.
To find the value of the random variable that has the least probability, we need to calculate the probability for each possible value of X.
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The probability that no white balls are transferred (X=0) is the probability of choosing 5 black balls from the 6 available in box A. This can be calculated using the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, and r is the number of items to choose.
So, P(X=0) = C(6, 5) / C(9, 5)
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The probability that one white ball is transferred (X=1) is the probability of choosing 1 white ball and 4 black balls.
So, P(X=1) = [C(3, 1) * C(6, 4)] / C(9, 5)
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The probability that two white balls are transferred (X=2) is the probability of choosing 2 white balls and 3 black balls.
So, P(X=2) = [C(3, 2) * C(6, 3)] / C(9, 5)
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The probability that all three white balls are transferred (X=3) is the probability of choosing 3 white balls and 2 black balls.
So, P(X=3) = [C(3, 3) * C(6, 2)] / C(9, 5)
After calculating these probabilities, the value of X that has the least probability is the one with the smallest probability value.
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