Maths (commerce) Solve the Following Question.(3 Marks)

2.Let $Y$ be the event that problem is not solved
$
\begin{aligned}
& \therefore P(Y)=P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right) \\
& =P\left(A^{\prime}\right) P\left(B^{\prime}\right) P\left(C^{\prime}\right) \\
& =\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \\
& =\frac{2}{5}
\end{aligned}
$

3.Let $Z$ be the event that exactly two students solve the problem.
$
\begin{aligned}
& \therefore P(Z)=P\left(A \cap B \cap C^{\prime}\right) \cup P\left(A \cap B^{\prime} \cap C\right) \cup P\left(A^{\prime} \cap B \cap C\right) \\
& =P(A) \cdot P(B) \cdot P\left(C^{\prime}\right)+P(A) \cdot P\left(B^{\prime}\right) \cdot P(C)+P\left(A^{\prime}\right) \cdot P(B) \cdot P(C) \\
& =\left(\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}\right)+\left(\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right) \\
& =\frac{4}{60}+\frac{3}{60}+\frac{2}{60} \\
& =\frac{3}{20}
\end{aligned}
$

There are 6 odd numbers $(1,3,5,7,9,11)$ and 5 even numbers $(2,4,6,8,10)$ from 1 to 11 .
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})=6 \times 6+5 \times 5 \\
& =36+25 \\
& =61 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{61}{121}
\end{aligned}
$
Let $B$ be the event that the numbers tickets drawn are odd
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{B})=6 \times 6=36 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}
\end{aligned}
$
Since 6 odd numbers are common between $A$ and $B$.
$
\begin{aligned}
& \therefore n(A \cap B)=6 \times 6=36 \\
& \therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{36}{121}
\end{aligned}
$
$\therefore$ Probability of both the numbers are odd, given that sum is even, is given by
$
P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{\frac{36}{121}}{\frac{61}{121}}=\frac{36}{61}
$