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 atleast once is
A
1−(3145)3
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B
1−(4345)3
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C
1−(4243)3
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D
1−(4145)3
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Solution
The correct option is B1−(4345)3 Given numbers are 10, 11, 12, 13, . . . , 99. The numbers with their digits product is 12 are 26, 34, 43, 62 ie, only 4. ∴P=P(E)=490=245q=P(¯E)=1−245=4345 Hence, the probability that he will laugh at least once =1−q3=1−(4345)3