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Let $X \sim \mbox{Binomial}(10, 0.2)$ and $Y \sim \mbox{Binomial}(10, 0.6)$. If we know that $Z = \min(X, Y)$, calculate $P(Z = 10)$.

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case 1: X = 10 and Y >= 10 case 2: X >= 10 and Y = 10 –  TenaliRaman Nov 11 '12 at 19:05
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1 Answer 1

Since in both cases you have 10 trials, and you need that the smaller number of successes be 10, thus you need successes in all trials. Then it is fairly easy, since the probability of getting succes is 0.2 and 0.6 respectively, thus giving you P$(Z=10)=0.2^{10}*0.6^{10}$

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