Question

Find the probability that:

Rolling a fair die 3 times, at least two dice are less than 4.

Previous answers have suggested 0.875 (189 feasible cases/ 216 total cases). However, from simple trial and error we can see that there are more than 27 cases where this answers fails. For example is the first dice gets a 4 and the 2nd dice gets 4,5,6 and the last dice gets any number, there is already 18 cases here that are not feasible.

Thank you.

Answer 1

Solution

At least two dice are less than 4 => 2 or more dice are less than 4

i.e., exactly 2 dice are less than 4 or all three dice are less than 4

Case 1:

Exactly 2 dice are less than 4 => 2 dice are less than 4 and the third die not less than 4.

Now, probability of less than 4 = 3/6 = ½. Probability of not less than 4 = 3/6 = ½

Thus for one combination of 2 dice are less than 4 and the third die not less than 4, the probability is

½ x ½ x ½ = 1/8.

There are 3 possible combinations of ‘2 dice are less than 4 and the third die not less than 4’.

So, finally, P(Exactly 2 dice are less than 4) = 3/8.

Case 2:

All 3 dice are less than 4

Probability as explained above is ½ x ½ x ½ = 1/8.

Since there is only one possible combination, the probability is just 1/8.

Thus, probability of at least two dice are less than 4 = 3/8 + 1/8 = ½ Answer

DONE

just a point

It is not at least two dice ....... We are not rolling 3 dice; we are only rolling one die but 3 times. So, it should be at least 2 rolls are less than 4 etc.