A fair die is thrown two times find the probability that sum of the numbers on them is 8

QUESTION 5

A fair die is thrown two times.

(a). Construct a table of the outcomes.

(b). Calculate the probability that the:

(i). sum of the outcomes is 8;

(ii). product of outcomes is less than 10;

(iii). outcomes contain at least a 3.

The Chief Examiner reported that this question was very popular among the candidates and their performance was commended. Majority of them were reported to obtain the following table of outcomes:

	1	2	3	4	5	6
1	1, 1	1, 2	1, 3	1, 4	1, 5	1, 6
2	2, 1	2, 2	2, 3	2, 4	2, 5	2, 6
3	3, 1	3, 2	3, 3	3, 4	3, 5	3, 6
4	4, 1	4, 2	4, 3	4, 4	4, 5	4, 6
5	5, 1	5, 2	5, 3	5, 4	5, 5	5, 6
6	6, 1	6, 2	6, 3	6, 4	6, 5	6, 6

From the table, total number of outcomes = 36, number of outcomes whose sum was 8 = 5. Therefore, probability that the sum of the outcomes was 8 =

. The Outcomes whose product was less than 10 = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1), (6, 1)}. Number of outcomes whose product was less than 10 = 17. Therefore, probability that product of outcome was less than 10 =

. Outcomes which contained at least a 3 = {(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6)}. Number of outcomes which contained at least a 3 = 11. Therefore, probability of at least a 3 =

Solution:

When two dice are thrown simultaneously, the sample space of the experiment is

{(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)}

So there are 36 equally likely outcomes.

Possible number of outcomes = 36.

(i)Let E be an event of getting a doublet.

Favourable outcomes = {(1,1), (2,2),(3,3), (4,4), (5,5),(6,6)}

Number of favourable outcomes = 6

P(E) = 6/36 = 1/6

Probability of getting a doublet is 1/6 .

(ii)Let E be an event of getting a sum of 8.

Favourable outcomes = {(2,6), (3,5), (4,4), (5,3), (6,2)}

Number of favourable outcomes = 5

P(E) = 5/36

Probability of getting a sum of 8 is 5/36.

When a fair die is tossed twice, the sample sp ace S is given by

S = {(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)}

∴ n (S) = 36

Let B = event that sum of numbers is at least 8

∴ B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

∴ n(B) = 15

∴ P(B) = `("n"("B"))/("n"("S"))`

= `15/36`

= `5/12`