Question

Two dice are thrown together, and the number appearing on them noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?

Answer 1

When two fair dice are thrown there are total 36 possible outcomes.

∵ X denotes the sum of 2 numbers appearing on dice.

∴ X can take values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

And also the appearance of numbers on two different dice is an independent event. So two find conditions like P(1 in the first dice and 2 in the second dice) can be given using multiplication rule of probability.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

P(X = 2) = \(\frac{1}{36}\) {∵ (1, 1) is the only combination resulting sum = 2}

{∵ (1, 2) and (2, 1) are the combinations resulting in sum = 3}

{∵ (1, 3), (3, 1) and (2, 2) are the combinations resulting in sum = 4}

{∵ (3, 2) (2, 3) (1, 4) and (4, 1) are the combinations resulting in sum = 5}

P(X = 6) = \(\frac{5}{36}\)

{∵ (1, 5) (5, 1) (2, 4) (4, 2) (3, 3) are the combinations resulting in sum = 6}

{∵ (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) are the combinations resulting in sum = 7}

P(X = 8) = \(\frac{5}{36}\)

{∵ (3, 5) (5, 3) (2, 6) (6, 2) (4, 4) are the combinations resulting in sum = 8}

P(X = 9) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

{∵ (3, 6) (6, 3) (5, 4) and (4, 5) are the combinations resulting in sum = 9}

P(X = 10) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

{∵ (6, 4), (4, 6) and (5, 5) are the combinations resulting in sum = 10}

P(X = 11) = \(\frac{2}{36}\) = \(\frac{1}{18}\)

{∵ (5, 6) and (6, 5) are the combinations resulting in sum = 11}

P(X = 12) = \(\frac{1}{36}\) {∵ (6, 6) is the only combination resulting sum = 2}

Now we have p_i and x_i.

∴ Required probability distribution is:-