A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Solution

Here n = 45
p = P (ball drawn is marked with 0) = $1 over 10$
$left parenthesis because space only space one space ball space out space of space 10 space is space marked space with space 0 right parenthesis$
$therefore space space space straight q space equals space 1 minus straight p space equals space 1 minus 1 over 10 space equals 9 over 10$
P(none is marked with the digit 0) = P(0) = $straight C presuperscript 4 subscript 0 space open parentheses 9 over 10 close parentheses to the power of 4$
$equals space 1 space cross times space open parentheses 9 over 10 close parentheses to the power of 4 space equals space open parentheses 9 over 10 close parentheses to the power of 4 space equals 6561 over 10000.$

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Probability

A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

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