+11 What is the probability of having a result of more than 12 if the die is rolled thrice, given the condition that the first roll is odd?
+118658 What is the probability of having a result of more than 12 if the die is rolled thrice, given the condition that the first roll is odd?
first throw 1 (12 more needed) 1 favourable outcomes
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | n | n | n | n | n | n |
| 2 | n | n | n | n | n | n |
| 3 | n | n | n | n | n | n |
| 4 | n | n | n | n | n | n |
| 5 | n | n | n | n | n | |
| 6 | n | n | n | n | Y |
first throw 3 (10 more needed) 6 favourable outcomes
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | ||||||
| 2 | ||||||
| 3 | ||||||
| 4 | Y | |||||
| 5 | Y | Y | ||||
| 6 | Y | Y | Y |
first throw 5 8 more needed 15 favourable outcomes
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | ||||||
| 2 | y | |||||
| 3 | y | y | ||||
| 4 | y | y | y | |||
| 5 | y | y | y | y | ||
| 6 | y | y | y | y | y |
total of posible throws is 3*36 = 108
favourable outcomes = 1+6+15 = 22
P(more than 12 ) = \(\frac{12}{108}=\frac{3}{27}=\frac{1}{9}\)
+129840 There are 3 possibilities if the first roll is odd. And with each of these, there are 36 possibilites = 3 * 36 = 108 possibilities in all......so we have
1, 6, 6
3, 6, 6
3, 6, 5
3, 5, 6
3, 5, 5
3, 4, 6
3, 6, 4
5, 6, 6 5, 4, 4
5, 6, 5 5, 6, 3
5, 5, 6 5, 3, 6
5, 5, 5 5, 5, 3
5, 4, 6 5, 3, 5
5 6, 4 5, 2, 6
5, 4, 5 5, 6, 2
5, 5, 4
And this is 22 ways.......so ....the probability = 22/108 = 11/54 ≈ 20.37%
