1) first person can choose anything number which translates to 4/4
2) second person choosing other number that hasn't been chosen 3/4
3) Third person choosing ...2/4
4) Last person = 1/4
multipy them 4*3*2*1/4*4*4*4 = 6/64
Please tell the OA.

Diitto aa2kash. 3/32

CrackGMAC wrote:Diitto aa2kash. 3/32

I beg to differ. 3/32 is less than 10%. Does that mean its more than 90 % chances that all 4 will pick the same number !!?

I agree with the 1 - (1/64) = 63/64 calculation.
What is the OA ?

Answer is 63/64. But is my approach correct?

crackgmat007 wrote:Answer is 63/64. But is my approach correct?

absolutely

crackgmat007 wrote:Answer is 63/64. But is my approach correct?

What is the source of this question??
If you are reffering to a standard material then definately you will see the Answer to be 3/32.

As per your method the answer is coming out to be 63/64 i.e nearly 99%. If you think logically then also its not possible to be that high. There is high probability that any 2 person choose the same number.

crackgmat007 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

Is this correct way to solve:

1. Probability that same number is selected by all 4 people

First number - irrelevant
Second, third, & fourth number - 1/4 each

Hence 1/4 * 1/4 * 1/4 = 1/64

2. To answer the question

1 - 1/64 = 63/64

As per your method you are first finding the probability of picking up the same number.
then you are subtracting it from 1. This approach would have been correct, if the question was formed like. "What is the approximate likelihood that all four people will not choose the same numbers?"

Also this 63/64 contains the cases where 2 people have choosen the same number & 3 people have chossen the same number.

I hope you agree with me now.

aa2kash wrote:

crackgmat007 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

Is this correct way to solve:

1. Probability that same number is selected by all 4 people

First number - irrelevant
Second, third, & fourth number - 1/4 each

Hence 1/4 * 1/4 * 1/4 = 1/64

2. To answer the question

1 - 1/64 = 63/64

As per your method you are first finding the probability of picking up the same number.
then you are subtracting it from 1. This approach would have been correct, if the question was formed like. "What is the approximate likelihood that all four people will not choose the same numbers?"

Also this 63/64 contains the cases where 2 people have choosen the same number & 3 people have chossen the same number.

I hope you agree with me now.

I understand what you are saying but these cases should be excluded when 2 people pick the same numbers. This is because now the All four will pick different numbers does not sound right anymore.

crackgmat007 wrote:Answer is 63/64. But is my approach correct?

in fact, it is not, because (1-everyone choose the same number) =not everyone choose the same number.
this means that 3 out of 4 can choose the same number, but not all four.

at least the fact that 63/64 is not equal to 3/32 shows us that this solution is not good.

actually we should think in this way-

4!/4^4