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

Here $straight n space equals space 5 comma space space space straight p space equals space 10 over 100 space equals 1 over 10 comma space space straight q space space equals 1 minus 1 over 10 space equals space 9 over 10$
P(all divisible by 10) = P(5) = $straight C presuperscript 5 subscript 5 space open parentheses 9 over 10 close parentheses to the power of 5 minus 5 end exponent space open parentheses 1 over 10 close parentheses to the power of 5 space equals space 1 space cross times space 1 space cross times space open parentheses 1 over 10 close parentheses to the power of 5 space equals space open parentheses 1 over 10 close parentheses to the power of 5$

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Probability

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.

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