Probability of sum 15 when roll 3rd dice, and first roll of 2 dice at least 10

Question

enter image description hereTwo dice have been thown, giving a total of at least 10. What is the probability that the throw of a third die will bring the total of the three numbers shown to 15 or higher? I work out is 2/27. but physicsandmathstutor.com give a solution is 4/9. i wander if you can help and explain in details. thanks

Answer 1

Hint: I suggest a Probability Tree with total values then if it is 15 or more as the second branch. Calculate the probabilities individually. This should help clear any doubt you have with your answer.

Answer 2

Let $A$ be the event "total of the three numbers shown to 15 or higher", $B_1$ be the event "total of first two is 10", $B_2$ be the event "total of first two is 11", $B_3$ be the event "total of first two is 12". We have \begin{align} P(B_1) = 3/6^2\\ P(B_2) = 2/6^2\\ P(B_3) = 1/6^2 \end{align} If $B$ is the event "total of first two is at least 10", then $P(B_1|B)=1/2$, $P(B_2|B)=1/3$ and $P(B_3|B)=1/6$. Finally \begin{align} P(A|B)&=P(A|B_1,B)P(B_1|B)+P(A|B_2,B)P(B_2|B)+P(A|B_3,B)P(B_3|B)\\ &=1/3\times1/2+1/2\times1/3+2/3\times1/6=4/9 \end{align}

Answer 3

if you list all possible events is 16, over sample space 216 events. 16/216 = 2/27

    D1   D2   D3
1   4 -> 6 => 5
2   4 -> 6 => 6
3   5 -> 5 => 5
4   5 -> 5 => 6
5   6 -> 4 => 5
6   6 -> 4 => 6
7   5 -> 6 => 4
8   5 -> 6 => 5
9   5 -> 6 => 6
10  6 -> 5 => 4
11  6 -> 5 => 5
12  6 -> 5 => 6
13  6 -> 6 => 3
14  6 -> 6 => 4
15  6 -> 6 => 5
16  6 -> 6 => 6

for people think use conditional probability, I think the wrong part is use probabilty 1/6 for two dice sum 10, applies to each event.