A number is chosen at random from the numbers 10 to 99. by seeing the number, a man will laugh if product of the digits is 12. If he chosen three numbers with replacement then the probability that he will laugh at least once is ............
A
1−(4345)3
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B
1−(245)3
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C
(4345)3
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D
(245)3
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Solution
The correct option is A1−(4345)3 The numbers between 10-99 whose product of the digits give 12 are 26,34,43,62−S Let P(X) denote the number of times the man laughs. P(X≥1) =1−P(X=0) Now P(x=0) implies that all the three numbers are not from S Hence P(X=0) =(8690)3 =(4345)3 Hence P(X≥1) =1−(4345)3