Probability of sum of 2 dice being greater than 10

There is a question of the conditional probability that say (That's not homework) : "What is the probability that the sum of 2 dice is greater or equals to 10 if :

1. we already know that the first dice is 5. Well if we know that the first dice is 5 the probability is: 2/6. well, I know how to compute this.
2. At least one of them is equal to 5. the solutions manual say that's 3/11 but I don't know how to get that.

In my head logically I would try this.

$V$ = "The sum of two dice is greater or equals than 10"

$S_1$ = "The first dice is a 5"

$S_2$ = "The second dice is a 5"

and then the probability result will be: $P(V \mid (S_1 \cup S_2))$ but this gives me the wrong result.

Solution 1:

Both of them can be $5$ as well. So if the event of at least one $5$ is $B$ then,

$ \displaystyle P(B) = 2 \cdot \frac{1}{6} \cdot \frac{5}{6} + \frac 16 \cdot \frac 16 = \frac{11}{36}$

Or using complimentary method, $ \displaystyle P(B) = 1 - \frac 56 \cdot \frac 56 = \frac{11}{36}$

Now if $A$ is the event of sum being $10$ or more,

$ \displaystyle P(A \cap B) = 2 \cdot \frac 16 \cdot \frac 16 + \frac 16 \cdot \frac 16 = \frac{3}{36}$

As there are $3$ favorable outcomes out of $36$: $\{5, 5\}, \{5, 6\}, \{6, 5\}$

So, $ \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} ~ $ does give you $\dfrac{3}{11}$

Solution 2:

The easiest way is to write down the entire sample space, first

1)

$$\Omega=\{(5,1);(5,2);(5,3);(5,4);(5,5);(5,6)\}$$

The answer is $2/6$ given that you have 2 favourable events among 6 possible

2): look at the following picture

yellow events are the possible ones while the red circle events are the favourable ones...thus the result is 3/11