duongthang wrote:38. As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive.
What is the approximate likelihood that 2 people will choose same number?
For this we have two cases to consider.
1. 2 ppl will select same and 2 ppl will select different digits [A,A,B,C]
Probability = [ (No. of ways to choose which 2 persons will have the same number) * (No. of ways to choose which number it will be) * (No. of ways to choose 2 different numbers out of 3 left ) * (No. of ways to choose which person it will be out of the left 2)] / [Total possibilities]
So, No. of ways to choose which 2 persons will have the same number = C(4,2) = 6
No. of ways to choose which number it will be = C(4,1) = 4
No. of ways to choose 2 different numbers out of 3 left = C(3,2) = 3
No. of ways to choose which person it will be = C(2,1) = 2
Total possibilities = 4^4 = 256
Thus, P1 = (6*4*3*2)/256 = 144/256
2. 2 ppl will select same and 2 ppl will select other two same digits [A,A,B,B]
Probability = [ (No. of ways to choose which 2 persons will have the same number) * (No. of ways to choose which number it will be) / [Total possibilities]
No. of ways to choose which 2 persons will have the same number = C(4,2) = 6
No. of ways to choose which number it will be = C(4,2) = 6
Thus, P2 = 6*6/256 = 36/256.
Required Answer = 144/256 + 36/256 = 180/256
duongthang wrote:What is the approximate likelihood that 3 people will choose same number?
Probability = [ (No. of ways to choose which 3 persons out of 4 will have same number) * (No. of ways the number will be selecetd) * (No. of ways last person will be selected)] / [Total possibilities]
No. of ways to choose which 3 persons out of 4 will have same number = C(4,3) = 4
No. of ways the number will be selecetd = C(4,1) = 4
No. of ways last person will be selected = 3
Answer = 3*4*4/256 = 48/256
