akshatgupta87 wrote:Can someone explain the logic behind this question.
Q) 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?
a> 9%
b> 12%
c> 16%
d> 20%
e> 25%
TIA
~Akshat
The first person can choose any of the 4 numbers.
P(2nd number is different) = 3/4. (Out of the 4 numbers, we can't select the first number chosen, leaving us 4-1=3 good options.)
P(3rd number is different) = 2/4. (Out of the 4 numbers, we can't select the first 2 numbers chosen, leaving us 4-2=2 good options.)
P(4th number is different) = 1/4. (Out of the 4 numbers, we can't select the first 3 numbers chosen, leaving us 4-3=1 good option.)
Since we want all of the events above to happen together, we multiply the fractions:
3/4 * 2/4 * 1/4 = 3/32 = 9/96 ≈ 9%.
The correct answer is
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