Question

Two dice are thrown. Find the probability of the following events.
3) The sum of the numbers on their upper faces is at least 8.
1) The sum of the numbers on their upper faces is at the most 4.​

Answer 1

Answer:

3) 4+4;6+2;5+3

1)4+1;3+2;4+2;5+1;6+1;3+5;6+2;4+5;6+4;5+5;5+6;6+6

Answer 2

3) Probability is $\frac{5}{12}$

1) Probability is $\frac{1}{6}$

Step-by-step explanation:

Two dice are thrown.

Sample space = 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)}

So, n(S) = 36

3) Let E be the event that the sum of the numbers on their upper face is at least 8.

E = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

So, n(E) = 15

Hence, Probability = $\frac{n(E)}{n(S)}$ = $\frac{15}{36} = \frac{5}{12}$

1) Let F be the event that the sum of the numbers on their upper face is at most 4.

F = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}

So, n(F) = 6

Hence, Probability = $\frac{n(F)}{n(S)} = \frac{6}{36} =\frac{1}{6}$