I know how to approach die problem when it is simple, However the given problem is kind of confusing.

Question

A die is thrown 3 times. We are asked to find the probability that points number of the first and second throw was 1, if points number of the third throw is greater than sum of first and second throw.

Answer 1

OK let's see. I think there are a couple of steps here. It helps to label the relevant random variables; note this would help in your asking of the question, too: So let $X_1$, $X_2$ and $X_3$ be the random variables corresponding to the outcomes of the dice throws. If I've understood correctly, we are looking for: $$ \mathbb{P}( X_1 = X_2 = 1 | X_3 > X_1 + X_2). $$ To be even more convenient, I'm going to write $Y = X_1 + X_2$. The problem doesn't distinguish between $X_1$ and $X_2$ and we are only ever interested in their sum (because $X_1 = X_2 = 1$ if and only if $Y=2$). So we are looking for $$ \mathbb{P}( Y = 2 | X_3 > Y). $$ Now by the usual Bayes' rule: $$ \mathbb{P}( Y = 2 | X_3 > Y) = \frac{\mathbb{P}(Y=2\ \text{and}\ X_3 > Y)}{\mathbb{P}(X_3 > Y)} = \frac{\mathbb{P}(X_3 > Y | Y = 2) \mathbb{P}(Y= 2)}{\mathbb{P}(X_3 > Y)}. $$ So to compute this, we will need to know $\mathbb{P}(X_3 > Y)$. By the law of total probability and the fact that we cannot have $X_3 > Y$ unless $Y \leq 5$, we have: \begin{align} \mathbb{P}(X_3 > Y) &= \sum_{j=2}^5 \mathbb{P}(X_3 > Y\ \text{and}\ Y = j) \\ &= \sum_{j=2}^5 \mathbb{P}(X_3 > Y | Y = j)\, \mathbb{P}(Y=j) \\ &= \frac{4}{6}\, \frac{1}{36} + \frac{3}{6}\, \frac{2}{36} + \frac{2}{6}\, \frac{3}{36} + \frac{1}{6}\, \frac{4}{36} \\ &= \frac{20}{216} \\ &= \frac{5}{54} \end{align} And the numerator in the expression we are trying to calculate is $$ \mathbb{P}(X_3 > Y | Y = 2) \mathbb{P}(Y= 2) = \frac{4}{6}\frac{1}{36} = \frac{1}{54} $$ So the answer is $$ \frac{1}{5}. $$