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# Please could someone help with the below statistics question please? watch

1. The discrete random variable x represents the score on the uppermost face of a biased die with six faces. The probability distribution of X is shown in the table below.
x 1 2 3 4 5 6
P(X= x) 0.1 0.1 0.1 0.2 0.2 0.3

The discrete random variable y represents the score on the uppermost face of a biased die with five faces. The probability distribution of Y is shown in the table below.

y 1 2 3 4 5
P(Y= y) 0.1 0.1 0.4 0.2 0.2

Each die is rolled once. The scores on the two dice are independent.
Find the probability that the sum of the two scores equals 2.

P(X=1)P(Y=1) = 0.1 x 0.1 = 0.01

and not P(X=1)P(Y=1) = 0.1 x 0.1 OR P(Y=1)P(X=1) = (0.1x0.1) + (0.1x0.1) = 0.02.

I don't understand why you don't account for the possibility that you choose the other die first.
2. It might help to think about this in terms of a tree. Let's say the first tier of the tree represents the die with six faces. This first tier will have six branches, and for each of those there will be five sub-branches to represent the die with five faces. You could take the dice in the opposite order - it makes no difference, you will end up with 30 sub-branches either way. Only one of the 30 sub-branches corresponds to a score of 1 on both dice. Order doesn't come into it - there is only one way to score two in total.

Where you would need to think about order is if you were looking at the probability of a total score of (say) 3. There would be two ways to do that: 2 on the 6-die and 1 on the 5-die or 1 on the 6-die and 2 on the 5-die. Then you would need to add the probabilities of scoring 2+1 and 1+2 (and those two outcomes would be represented by two separate sub-branches of the tree mentioned above).
3. Thanks so much - that's a really helpful explanation!

(Original post by old_engineer)
It might help to think about this in terms of a tree. Let's say the first tier of the tree represents the die with six faces. This first tier will have six branches, and for each of those there will be five sub-branches to represent the die with five faces. You could take the dice in the opposite order - it makes no difference, you will end up with 30 sub-branches either way. Only one of the 30 sub-branches corresponds to a score of 1 on both dice. Order doesn't come into it - there is only one way to score two in total.

Where you would need to think about order is if you were looking at the probability of a total score of (say) 3. There would be two ways to do that: 2 on the 6-die and 1 on the 5-die or 1 on the 6-die and 2 on the 5-die. Then you would need to add the probabilities of scoring 2+1 and 1+2 (and those two outcomes would be represented by two separate sub-branches of the tree mentioned above).

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