A six-faced unbiased die is thrown twice and the sum of the numbers appearing on the upper face is observed…
A six-faced unbiased die is thrown twice and the sum of the numbers appearing on the upper face is observed to be 7 . The probability that the number 3 has appeared atleast once, is
$\frac{1}{5}$
$\frac{2}{5}$
$\frac{3}{5}$
$\frac{4}{5}$
Solution
Sum of the dice is 7 .
$
\begin{aligned}
& S=-\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\} \\
& \therefore n(S)=6
\end{aligned}
$
Let, $E=$ Event of getting atleast three 3 or a die
$
\therefore n(E)=2
$
$
\therefore \text { Required probability }=\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3}
$