Two fair dice are tossed together once.
(a) Draw a sample space for the possible outcomes ;
(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.
(a) Draw a sample space for the possible outcomes ;
(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.
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Correct Answer: Option n
Explanation:
(a)
(b) (i) P(getting a total of 7) = \(\frac{6}{36}\)
P(getting a total of 8) = \(\frac{5}{36}\)
P(getting a total of 7 or 8) = \(\frac{6}{36} + \frac{5}{36}\)
= \(\frac{11}{36}\)
(ii) P(getting a total less than 4) = \(\frac{3}{36}\)
= \(\frac{1}{12}\)
(a)
| 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 |
(b) (i) P(getting a total of 7) = \(\frac{6}{36}\)
P(getting a total of 8) = \(\frac{5}{36}\)
P(getting a total of 7 or 8) = \(\frac{6}{36} + \frac{5}{36}\)
= \(\frac{11}{36}\)
(ii) P(getting a total less than 4) = \(\frac{3}{36}\)
= \(\frac{1}{12}\)