Rolling a die $4$ times. Probability of getting a (3) once and a (6) twice.

Question

I am stuck on this question and don't have an answer key, so I can't check or see any method that leads to the solution.

My thoughts: Add the probability of getting a (3) to the probability of getting two (6). But this can't be correct can it?

Cheers and thank you in advance.

Answer 1

The strategy stated below is based on the assumption that we wish to find the probability that exactly one three and exactly two sixes occur in four rolls of the die. Otherwise, we would have to add the probabilities of two threes and two sixes and of one three and three sixes to the numerator.

Strategy

1. For the denominator, how many sequences of outcomes are there when a die is rolled four times?
2. For the numerator, choose two positions for the sixes, choose one of the remaining positions for the three, and choose which of the other numbers fills the remaining open position.

Note that you only add the probabilities of mutually exclusive events, that is, events that cannot occur simultaneously. However, it is possible to roll a three once and a six twice in four rolls of a die, so your approach will not work. When events can occur simultaneously, the probabilities multiply.

Probability that there is exactly one three and exactly two sixes in four rolls of a fair die: There are six possible outcomes for each of the four rolls, so there are $6^4$ outcomes in our sample space. For the favorable cases, choose two of the four positions for the sixes, one of the remaining two positions for the three, and choose one of the other four outcomes for the remaining open position. Hence, there are $$\binom{4}{2}\binom{2}{1}\binom{4}{1}$$ favorable outcomes. Therefore, the probability of exactly one three and exactly two sixes in four rolls of a fair die is

$$\frac{\dbinom{4}{2}\dbinom{2}{1}\dbinom{4}{1}}{6^4}$$

Probability of at least one three and at least two sixes in four rolls of a fair die: To the favorable cases above, we must add the cases two threes and two sixes and one three and three sixes, which gives

$$\frac{\dbinom{4}{2}\dbinom{2}{1}\dbinom{4}{1} + \dbinom{4}{2} + \dbinom{4}{1}}{6^4}$$

For the second term in the numerator, the number of ways two threes and two sixes can occur is determined by choosing which two positions in the sequence of four rolls the threes occupy since the sixes must occupy the remaining two positions. Similarly, for the third term in the numerator, the number of ways one three and three sixes can occur is determined by choosing the position of the three in the sequence of four rolls since the sixes must occupy the remaining three positions.

Answer 2

Based on what N.F.Taussig said i would say:

• There (3,6,6,a) where a is a number between 1 and 6
• There are 6 Values that a can assume
• The order of the first 3 Numbers can be rearranged. So i would say that there are 3 different ways in which (3,6,6) can be arranged.

so the probability would be (3*6)/1296

So 18/1296 ?

My apologies for the terrible formatting. I dont really know how to insert Fractions and Mathematical formulas.