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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
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Verified Answer
The correct answer is:
$\frac{3}{5}$
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
$$
\therefore n(E)=2
$$
$$
\therefore \text { Required probability }=\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3}
$$
$$
\begin{aligned}
& S=-\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\} \\
& \therefore n(S)=6
\end{aligned}
$$
Let, $E=$ Event of getting atleast three 3 or a die
$$
\therefore n(E)=2
$$
$$
\therefore \text { Required probability }=\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3}
$$
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