In: Math
Two fair dice are rolled and the outcomes are recorded. Let X denotes the larger of the two numbers obtained and Y the smaller of the two numbers obtained. Determine probability mass functions for X and Y, and the cumulative distribution functions for X and for Y. Present the two cumulative distribution functions in a plot. Calculate E (2X + 2Y −8).
Let X denotes the larger of the two numbers obtained . So the possible values of X are 1, 2, 3, 4, 5, and 6 with frequencies as 1, 3, 5, 7, 9, and 11 respectively.
So the probability distribution function ( f(x) ) of X and cumulative distribution function ( F(x) ) are as follows:
X | 1 | 2 | 3 | 4 | 5 | 6 | Total |
P(X =x) | 1/36 | 3/36 | 5/36 | 7/36 | 9/36 | 11/36 | 1 |
F(x) | 1/36 | 4/36 | 9/36 | 16/36 | 25/36 | 1 |
Let Y denotes the smaller of the two numbers obtained . So the possible values of Y are 1, 2, 3, 4, 5, and 6 with frequencies as 11, 9, 7, 5, 3, and 1 respectively.
So the probability distribution function ( f(y) ) of Y and the cumulative distribution function ( F(y) ) of Y are as follows:
Y | 1 | 2 | 3 | 4 | 5 | 6 | Total |
P(Y = y) | 11/36 | 9/36 | 7/36 | 5/36 | 3/36 | 1/36 | 1 |
F(y) | 11/36 | 20/36 | 27/36 | 32/36 | 35/36 | 1 |
Let's find E(X) and E(Y):
E(X) = 1*(1/36) + 2* (3/36) + 3*(5/36) + 4*(7/36) + 5*(9/36) + 6*(11/36) = 161/36 = 4.4722
E(Y) = 1*(11/36) + 2*(9/36) + 3*(7/36) + 4*(5/36) + 5*(3/36) + 6*(1/36) = 91/36 = 2.5278
E (2X + 2Y −8) = 2*(161/36) + 2*(91/36) - 8 = 14 - 8 = 6