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:-

∴ Mean =

Variance =

Standard deviation = √variance =