Exam P Practice Problem 11 – Binomial Sums

Problem 11A
Suppose X and Y are independent binomial variables with the following probability functions:

    \displaystyle P(X=k)=\binom{5}{k} \ \biggl[\frac{6}{10}\biggr]^k \ \biggl[\frac{4}{10}\biggr]^{10-k} \text{ where } k=0.1,2,3,4,5

    \displaystyle P(Y=k)=\binom{5}{k} \ \biggl[\frac{6}{10}\biggr]^k \ \biggl[\frac{4}{10}\biggr]^{10-k} \text{ where } k=0.1,2,3,4,5

Calculate the following sum:

    \displaystyle P(X=0) \ P(Y=4)+P(X=1) \ P(Y=3)+P(X=2) \ P(Y=2)

      \displaystyle +P(X=3) \ P(Y=1) + P(X=4) \ P(Y=0)

Problem 11B
Suppose X and Y are independent binomial variables with the following probability functions:

    \displaystyle P(X=k)=\binom{6}{k} \ \biggl[\frac{1}{2}\biggr]^6 \text{ where } k=0.1,2,3,4,5,6

    \displaystyle P(Y=k)=\binom{4}{k} \ \biggl[\frac{1}{2}\biggr]^4 \text{ where } k=0.1,2,3,4

Calculate the following sum:

    \displaystyle P(X=1) \ P(Y=4)+P(X=2) \ P(Y=3)+P(X=3) \ P(Y=2)

      \displaystyle +P(X=4) \ P(Y=1) + P(X=5) \ P(Y=0)

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Answers

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Hint

Performing the calculation as the problem is stated is a trap. Falling for this trap could be costly. Note that the sum of the two random variables is also a binomial distribution and that the required calculation is merely a probability calculation of the sum. To see this, write out the probability function of the independent sum using the convolution method.

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2 responses

  1. I would like to see the solution for 11-a.

    1. For 11-a I’m getting [(5C0)(5C4)+(5C1)(5C3)+(5C2)(5C2)+(5C3)(5C1)+(5C4)(5C0)]*(0.6^4)(0.4^16)… which differs from your answer…

      And I can’t see any other way around calculating it this way.

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