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Re: List A = 0, 1, 2, 3, 4, 6 List B = 3, 6, 8, 9 Two numbers are randomly [#permalink]
Carcass wrote:
For the denominator, you have to pick 1 number among 4 and 1 among 6

For the numerato in this kind of question is just a matter of counting manually. It is even faster instead to stay there and figure it out with a formula

List A = 0, 1, 2, 3, 4, 6
List B = 3, 6, 8, 9

1) start from 0
3-6-8-9 all divisible by 9 once multiplied

2) with 1 we have only 1 case: 9

3) with 2 only 1:18

4) with 3 only 3:9,18,27

5) with 9 only 1:9

6) with 6 we have 3: 18,36,54

Totally 13

13/6*4=13/24

C



Yes i applied this method, but the total cases doesn't add up to 13.

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List A = 0, 1, 2, 3, 4, 6 List B = 3, 6, 8, 9 Two numbers are randomly [#permalink]
1
Expert Reply
Made a small editing . see now sir.

They are 13
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Re: List A = 0, 1, 2, 3, 4, 6 List B = 3, 6, 8, 9 Two numbers are randomly [#permalink]
1
case 1 : Multiple of 9 is always divisible by 9 thus we can select any number from all 6 numbers of list A and only 1 number that is 9 from list B
= 6C1 * 1C1 = 6

Case 2 : We know Zero is multiple of every number thus we can select only 1 number that is 0 from list A and any from all 4 numbers of list B
= 1C1 * 4C1 = 3

Case 3 : (3,3) (3,6) (6,6) (Rest cases which can't be taken through combination) = 3

thus total = 6+4+3 = 13

Total case = 6C1 * 4C1 = 24

What is the probability that the product of these two numbers is divisible by 9 = 13/24
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