Question

If a pair of regular dice are tossed once, use the expectation formula to determine the expected sum of the numbers on the upward faces of the two dice.

Answer 1

Answer:

The expected sum of the numbers on the upward faces of the two dice is 7.

Step-by-step explanation:

Consider the provided information.

If two pair of dice tossed the possible out comes are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),  (5,6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5),  (6,6)

Now we need to find the expected sum of the numbers on the upward faces of the two dice.

The expected sums can be:

Sum:    2,      3,       4,       5,      6,      7,       8,        9,      10,    11,      12

Prob: 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36

As we know that the expectation of experiment can be calculated as:

$P(S_1)\cdot S_1+P(S_2)\cdot S_2+........+P(S_n)\cdot S_n$

Here S represents the numerical outcomes and P(S) is the respective probability.

Substitute the respective values in the above formula.

$=2\times\frac{1}{36}+3\times\frac{2}{36}+4\times\frac{3}{36}+5\times\frac{4}{36}+6\times\frac{5}{36}+7\times\frac{6}{36}+8\times\frac{5}{36}+9\times\frac{4}{36}+10\times\frac{3}{36}+11\times\frac{2}{36}+12\times\frac{1}{36}\\=\frac{252}{36}\\=7$

Hence, the expected sum of the numbers on the upward faces of the two dice is 7.