Probability Prob

[garuhape]

If any number from set A is multiplied by any number from set B, what is the probability that the product is a multiple of 4?
A = {21, 22, 23, 24, 25} B = {23, 24, 25, 26, 27}

The OA is 40%. However mine is 36% because I only get 9 combination which are dividable by 4 and not 10!

Can someone help please!

[force5]

Hi Garuhape
this is how i did

Probability = required cases/total cases

total cases = 25
required cases:
if we choose 21 from A we can choose only 24 from B. -----1
22 from A . 24 and 26 (From B) ---------2
23 from A 24 only (from B)-------------1
25 from A 24 only (from B)-----------1
24 from A all the 5 (from B)-------------5

Hence reqd prob is 10/25 = 40%

hope that helps

[garuhape]

Hi Garuhape
this is how i did

Probability = required cases/total cases

total cases = 25
required cases:
if we choose 21 from A we can choose only 24 from B. -----1
22 from A . 24 and 26 (From B) ---------2
23 from A 24 only (from B)-------------1
25 from A 24 only (from B)-----------1
24 from A all the 5 (from B)-------------5

Hence reqd prob is 10/25 = 40%

hope that helps

I missed the 22*26 one. How do you know without too much calculation that the product dividable by 4? Did you multiply the 22 with each of the other integers and divided the products by 4?

[force5]

Good question garuhape.. For this i have to get back to the basics of prime factorization. any number to be a multiple of a number needs to ATLEAST have all the prime factors in its prime box. For example. 10 ( factors 2,5) cannot be a multiple of 4 (factors 2,2) since the prime box of 10 doesnt have all the factors of 4. on the other hand 12 ( factors 2,2,3) has all the prime factors of 4 (2,2) hence its a multiple of 4. The thing to notice here is that the prime box of the multiple may have other factors too( 12 has a factor 3 which is not a part of 4's prime box) so in a way a multiple is a super set of a smaller set Factor.

now In this question we need multiple of 4 which means that we want to search for the prime factors of 4 ( 2,2). hence we need 2 two's, Now you just need to locate even numbers ( every even number will have atleast one 2). No calculation is needed here. 6 multiplied by 22 will always be a multiple of 4 WHY( 6 has one 2 and 22 has one 2. so the prime box of 6*22 will always have 2 Two's).........

I Hope i tried to clear your problem.

[garuhape]

Thanks for this nice explanation:

22=2*11 and 26=2*13 ->26*22=2*2*11*13

[fr743]

Hi Garuhape
this is how i did

Probability = required cases/total cases

total cases = 25
required cases:
if we choose 21 from A we can choose only 24 from B. -----1
22 from A . 24 and 26 (From B) ---------2
23 from A 24 only (from B)-------------1
25 from A 24 only (from B)-----------1
24 from A all the 5 (from B)-------------5

Hence reqd prob is 10/25 = 40%

hope that helps

There are 3 two's in 24 (2*2*2*3). So, if you multiply two 24's you get 6 two's instead of 5 like you said. Could you explain me this, please?

[force5]

There are 3 two's in 24 (2*2*2*3). So, if you multiply two 24's you get 6 two's instead of 5 like you said. Could you explain me this, please?

what we want is to see if a*b is a multiple of 4 (2,2) so the number (multiple) should have prime factors of 4(2,2). when we multiply 24*24 then we have 6 twos, which means that it has all the prime factors of 4(2,2) . As i mentioned in the explanation that the multiple may have other factors too. But that doesn't concern us. It should at least have all the prime factors of the FACTOR.

i hope you are not mixing the two concepts... If you want to know why there are 5 choices when 24 is chosen in set A there here is the answer for that

if we choose 24 from set A then from B we can choose any number and the product will be divisible by 4. Hence if i choose 24 from A then from B i can choose (23,24,25,26,27) ie- 5 choices......