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A number is chosen at random from the number 10to99. By seeing the number a man will laugh if product of the digits is 12. If he choose three number with replacement then the probability that he will laugh at least once is

A

`1-((3)/(5))^(3)`

B

`(2(45^(2)+43^(2)-45xx43))/(45^(3)`

C

`(98)/(125)`

D

`1-((43)/(45))^(3)`

Text Solution

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The correct Answer is:
D

There can be four such numbers i.e 43,34,62,26.
Whose product of digit is 12
implies Probability that the man will laugh by seeing the chosen numbers `=(4)/(90)=(2)/(45)`
implies required probability
`=1-(1-(2)/(45))^(3)=1-((43)/45)^(3)`
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