Question

Given the digits 1,3,6,9 find the probability that a 3 digit number formed by using
them with no digit repeated is divisible by 4​

Answer 1

Answer:

Given the digits 1, 3, 6, 9 find the probability that a 3 digit number formed by using them with no digit repeated is divisible by 4?

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The total number of 3 digit numbers which can be formed by 4 digits with no digit repeated is equal to 4P3 which is equal to 4!1! where the unit digit can be occupied by any four of the digits the tens place by any 3 of the remaining digits and the 100th place with any of the remaining two digits while the last digit isn't taken into account. So total ways in which the numbers can be formed is 4×3×2 .

Now we need to take the favourable cases. Numbers which can be divided by 4 have the last 2 numbers divisible by 4. Thus the unit place digit must be even. So it can be filled in 1 way only. The tens place can be filled by any of the 3 digits 1,3 or 9 as 16, 36 and 96 are all divisible by 4. Thus till now total ways of forming a favourable number is 3×1 . Taking the 100th place digit too into account which can be filled by any of remaining 2 digits. We get the final answer as 1×3×2

So taking the probability by dividing favourable cases by unfacourable cases we get

1×3×24×3×2

Which on solving gives 1/4 as the final answer.

Answer 2

Answer:

the numbers can b 16,36,96,so unit digit can be filled in 1 way wheareas 10's place can be filled in 3 ways and hundred's place in 2 ways,so 2x3x1/4!=6/24=1/6

Step-by-step explanation:

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