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If a die is rolled twice and the sum of the numbers appearing on them is observed to be 6 , then the probability that the number 1 appears atleast once on them is
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Verified Answer
The correct answer is:
$\frac{2}{5}$
Consider the events
$A=$ Sum of the number appearing on them is 6 .
$B=$ Number 1 appears atleast once
$$
\begin{aligned}
& P(A)=\frac{5}{36} P(B)=\frac{11}{36} \\
& P(A \cap B)=\frac{2}{36} \\
&
\end{aligned}
$$
$\therefore$ Required probability $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{2 / 36}{5 / 36}=\frac{2}{5}$
$A=$ Sum of the number appearing on them is 6 .
$B=$ Number 1 appears atleast once
$$
\begin{aligned}
& P(A)=\frac{5}{36} P(B)=\frac{11}{36} \\
& P(A \cap B)=\frac{2}{36} \\
&
\end{aligned}
$$
$\therefore$ Required probability $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{2 / 36}{5 / 36}=\frac{2}{5}$
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