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)`

4 views

1 Answers

Correct Answer - 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)`

4 views

Related Questions