Question

A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Answer 1

When a pair of fair dice is thrown there are total 36 possible outcomes.

X denotes the minimum of two numbers which appear

∴ X can take values 1, 2, 3, 4, 5 and 6

P(X = 1) = 11/36

[Possible Pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)]

P(X = 2) = 9/36

[Possible Pairs: (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 6),(2, 5), (2, 4), (2, 3)]

P(X = 3) = 7/36

[Possible Pairs: (3, 3),(3, 4),(4, 3),(5, 3),(3, 5),(3, 6),(6, 3)]

P(X = 4) = 5/36

[Possible Pairs: (4, 4), (5, 4), (4, 5), (4, 6), (6, 4)]

P(X = 5) = 3/36

[Possible Pairs: (5, 5),(5, 6),(6, 5)]

P(X = 6) = 1/36

[Possible Pairs: (6, 6)]

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑x_i²p_i – (∑x_ip_i)²

Standard Deviation is given by SD = √ Variance

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

Required Probability distribution table:-

Standard deviation = √variance