Question

Suppose we toss two fair dice. Let X = i+j, where i is the outcome of the first die, and j is the outcome of the second die, with i, j ∈ {1, 2, 3, 4, 5, 6}.
Let p(x) = P(X=x) be the probability mass function of X. Round all answers to 4 decimal places.

p(2) =

Tries 0/5

p(3) =

Tries 0/5

p(4) =

Tries 0/5

p(5) =

Tries 0/5

p(6) =

Tries 0/5

p(7) =

Tries 0/5

p(8) =

Tries 0/5

p(9) =

Tries 0/5

p(10) =

Tries 0/5

p(11) =

Tries 0/5

p(12) =

Answer 1

Given:

Suppose we toss two fair dice. So the total number of outcomes = 36

Let X = i+j,

where i is the outcome of the first die, and j  is the outcome of the second die, with i, j ∈ {1, 2, 3, 4, 5, 6}.

Here we break the sum into each of every probable cases.

We find the corresponding probability by finding the total number of outcomes that gives the given sum of the die.

Let p(x) = P(X=x) be the probability mass function of X.

Now,

Probability , p(x) = number of favourable outcomes / Total number of outcomes

Therefore 1) p(2) = 0.0278

2) p(3) = 0.0556

3) p(4) = 0.0833

4) p(5) = 0.1111

5) p(6) = 0.1389

6) p(7) = 0.1667

7) p(8) = 0.1389

8) p(9) = 0.1111

9) p(10) = 0.0833

10) p(11) = 0.0556

11) p(12) = 0.0278