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 =
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`