pradeepkaushal9518 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?



9%



12%



16%



20%



25%

i arrived at[spoiler] 3/32, that is close to 9% [/spoiler] please post the OA

Same post has been discussed earlier here:
https://www.beatthegmat.com/probability-t46511.html

i will post oa after few discussions

Tricky question..

The first person can choose anything..So chance of success 1/1
Second Person can choose any three of the four numbers So chance of success 3/4
Third Person can choose any two of the four numbers So chance of success 2/4
Last Person can choose only one of the four numbers So chance of success 1/4

Hence we get, 1 * 3/4 * 2/4 * 1/4 = 6/64 = approx 9%

Pick A

There is an easily misguided way here..
Let me explain that also..
We will easily be trapped to find out "everyone chooses different number" and then subtract from 1.
In that case we have,
Probabaility for first person is 1
Second is 1/4, third 1/4, fourth 1/4
1/4 * 1/4* 1/4 = 1/64
1-1/64 = 63/64.

But the trap here is that "everyone chooses different number" is different from "all four people will choose different numbers"

Suppose the first person picked "3". Second person and third person also can pick "3". But the fourth one should not.

I hope I am clear.. Let me know if you have any doubt!!

yea 9% for me too, just like kvcp, altho i did not understand his trap, and the answer choices do not contain the trap answer, so its not much of a trap, is it?