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 choose three numbers with replacement then the probability that he will laugh at least once is
A
1−(35)3
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B
(4345)3
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C
1−(425)3
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D
1−(4345)3
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Solution
The correct option is D1−(4345)3 There can be four such numbers, i.e., 43,34,62,26
whose product of digit is 12.
Hence, Probability that a man will laugh by seeing the chosen numbers = 490=245
∴, the required probability = 1−(1−245)3=1−(4345)3