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Question
It is known that 60% of mice inoculated with a serum are protected from a certain disease. If 5 mice are inoculated, find the probability that more than 3 contract the disease .
Solution
Let X be the number of mice that contract the disease .
Then, X follows a binomial distribution with n =5.
Let p be the probability of mice that contract the disease.
\[\therefore p = 0 . 4 \text{ and } q = 0 . 6\]
\[\text{ Hence, the distribution is given by} \]
\[P(X = r) = ^{5}{}{C}_r \left( 0 . 4 \right)^r \left( 0 . 6 \right)^{5 - r} , r = 0, 1, 2, 3, 4, 5\]
\[ P(X > 3) = P(X = 4) + P(X = 5)\]
\[ = ^{5}{}{C}_4 \left( 0 . 4 \right)^4 \left( 0 . 6 \right)^{5 - 4} +^{5}{}{C}_5 \left( 0 . 4 \right)^5 \left( 0 . 6 \right)^{5 - 5} \]
\[ = 0 . 0768 + 0 . 01024\]
\[ = 0 . 08704\]
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