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Question
A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.
Solution
Let X be the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.
Then, X follows a binomial distribution with n =5;
\[p = \text{ Probability of getting a ticket bearing number divisible by } 10 . \]
\[ p = \frac{1}{100}(10) = \frac{1}{10}; q = \frac{9}{10}; \]
\[P(X = r) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r} \]
\[\text{ Probability that all thetickets bear numbers divisible by } 10\]
\[ = P(X = 5) = ^{5}{}{C}_5 \left( \frac{1}{10} \right)^5 \left( \frac{9}{10} \right)^{5 - 5} = \left( \frac{1}{10} \right)^5 \left( \frac{9}{10} \right)^0 = \left( \frac{1}{10} \right)^5 \]
Hence, required probability is \[\left( \frac{1}{10} \right)^5\]
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