Question

Q15. 4 men throw a die each simultaneously. find the probability that at least 2 people get the same number.

a. 5/18

b. 13/18

c. 1/36

d. 1/2

Answer 1

The first person rolls a die, and gets a number.

The second person has a 5/6 chance of rolling a different number.

The third person has a 4/6 chance of rolling a number different from those two.

The fourth person has a 3/6 chance of rolling a number different from all three.

The odds that no two will roll the same number is 5/6*4/6*3/6 = 5/18 = .2777...

That's the odds that no two will roll the same number, so the odds that two (or more) of them will roll the same number is 1-5/18 = 13/18 = .7222... is the answer

OR

One way to solve it is by considering the opposite - That on throwing the dice 4 times, you never get a pair.

In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are 5/6, 4/6 and 3/6 respectively.

So on multiplying we get 6/6*5/6*4/6*3/6 = 5/18

On substracting it from 1, we get 1-(5/18) = 13/18

Therefore 13/18 is the answer