Thank you for reporting, we will resolve it shortly
Q.
A die is thrown three times and the sum of three numbers obtained is $15$. The probability of first throw being $4$, is
Probability
Solution:
If first throw is four, then sum of numbers appearing on last two throws must be equal to eleven. That means last two throws are $(6,5)$ or $(5,6)$
There is a set having sum of numbers as $15$ is
$\{(3, 6, 6), (6, 3, 6), (6, 6, 3), (4, 6, 5), (4, 5, 6), (6, 5, 4)$,
$ (6,4, 5), (5,4, 6), (5,6,4), (5, 5, 5)\} = 10$.
Now, there are $10$ ways to get the sum as $15$.
$\Rightarrow $ Required probability $= \frac{2}{10} $
$= \frac{1}{5}$.