2. Suppose that Y is a binomial random variable based on n trials with success probability p and consider Y*=n-Y. a Argue that for y* = 0, 1,..., n P(Y* = y*) = P(n − Y = y*) = P(Y = n − y*). b Use the result from part (a) to show that n P(Y* = y*) = = (₁ " y₁) p^-' q²" = (".)q" p²-x. n-y* c The result in part (b) implies that Y* has a binomial distribution based on n trials and "success" probability p* = q = 1 - p. Why is this result "obvious"?

A First Course in Probability (10th Edition)
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ISBN:9780134753119
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Chapter1: Combinatorial Analysis
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2.
Suppose that Y is a binomial random variable based on n trials with success probability p and
consider Y* = n - Y.
a Argue that for y* = 0, 1, ..., n
P(Y* = y*) = P(n − Y = y*) = P(Y = n − y*).
b Use the result from part (a) to show that
n
P(Y* = y*) = · ( ₁₁² y.)pª²-xq" = ("₁. )¶" p²-x.
- y*
c The result in part (b) implies that Y* has a binomial distribution based on n trials and
"success" probability p* = q = 1 – p. Why is this result "obvious"?
Transcribed Image Text:2. Suppose that Y is a binomial random variable based on n trials with success probability p and consider Y* = n - Y. a Argue that for y* = 0, 1, ..., n P(Y* = y*) = P(n − Y = y*) = P(Y = n − y*). b Use the result from part (a) to show that n P(Y* = y*) = · ( ₁₁² y.)pª²-xq" = ("₁. )¶" p²-x. - y* c The result in part (b) implies that Y* has a binomial distribution based on n trials and "success" probability p* = q = 1 – p. Why is this result "obvious"?
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