+118725 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 ) = \(\frac{12}{108}=\frac{3}{27}=\frac{1}{9}\)
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