If one threedigit number 0 cannot be a left digit is chosen

If one three-digit number (0 cannot be a left digit) is chosen at random from all those that can be made from the following set of digits, find the probability that the one chosen is not a multiple of 5. The probability that the one chosen is not a multiple of 5 is

Solution

as it doesn\'t say anything about repeatition so i will assume repeatition of digits is allowed.

there are total 8 digits given {0,1,2,3,4,5,6,7}

first digit from left can\'t be 0 as aseked so there are 7 ways to fill it.

second digit can by any of them so there are 8 ways to fill it.

third digit can by any of them so there are 8 ways to fill it.

Hence total number of possible events =7*8*8=448

Now we count favorable number of events.

first and second digits will be filled in same ways

number is not a multiple of 5 means third digit can\'t be 0 or 5 so there are 6 ways to fill third digit

hence number of favorable events=7*8*6=336

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Hence required probability =(favorable no of events)/(total events)=336/448=3/4