Complete the following activities and rewrite it : (2M)

Question 12 Marks

If $A$ and $B$ are two independent events and $P(A)=\frac{3}{5}, P(B)=\frac{2}{3}$, find
i) $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ ii) $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$ iii) $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$ iv) $P\left(A^{\prime} \cap B^{\prime}\right)$ v) $P(A \cup B)$

Answer

$\mathrm{P}(\mathrm{A})=\frac{3}{5} \therefore \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=\frac{2}{5}$ $\mathrm{P}(\mathrm{B})=\frac{2}{3} \therefore \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=\frac{1}{3}$
i) $P(A \cap B)=P(A) P(B)=\frac{2}{5}$
ii) $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{1}{5}$
iii) $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}(\mathrm{B})=\frac{4}{15}$
iv) $P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) P\left(B^{\prime}\right)=\frac{2}{15}$
v) $P(A \cup B)=P(A)+P(B)-P(A \cap B)=\frac{13}{15}$

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Question 22 Marks

Two dice are thrown together. What is the probability that,
i) Sum of the numbers is divisible by 3 or 4?
ii) Sum of the numbers is neither divisible by 3 nor 5?

Answer

Let $\mathrm{S}$ be the sample space
$
\begin{aligned}
& \text { Let } \mathrm{N}_1=\mathrm{N}_2=\{1,2,3,4,5,6\} \\
& \mathrm{S}=\mathrm{N}_1 \times \mathrm{N}_2=\left\{(x, y) / x \in \mathrm{N}_1, y \in \mathrm{N}_2\right\} \\
& \mathrm{n}(\mathrm{S})=36
\end{aligned}
$
i) Let event A: Sum of the numbers is divisible by 3
$\therefore$ possible sums are $3,6,9,12$.
$
\begin{aligned}
\therefore & A=\{(1,2),(1,5),(2,1),(2,4),(3,3), \\
& (3,6),(4,2),(4,5),(5,1),(5,4),(6,3), \\
& (6,6)\} \\
\therefore & n(A)=12 \therefore P(A)=n(A) / n(S)=12 / 36
\end{aligned}
$
Let event B: Sum of the numbers is divisible by 4 .
$\therefore$ possible sums are $4,8,12$
$
\begin{aligned}
\therefore & \mathrm{B}=\{(1,3),(2,2),(2,6),(3,1),(3,5), \\
& (4,4),(5,3),(6,2),(6,6)\} \\
\therefore & \mathrm{n}(\mathrm{B})=9 \therefore \mathrm{P}(\mathrm{B})=\mathrm{n}(\mathrm{B}) / \mathrm{n}(\mathrm{S})=\frac{9}{36}
\end{aligned}
$
$\therefore$ Event $\mathrm{A} \cap \mathrm{B}$ : Sum of the numbers is divisible by 3 and 4 i.e. divisible by 12 .
$\therefore$ possible Sum is 12
$
\begin{aligned}
& \therefore \mathrm{A} \cap \mathrm{B}=\{(6,6)\} \\
& \therefore \mathrm{n}(\mathrm{A} \cap \mathrm{B})=1 \\
& \therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) / \mathrm{n}(\mathrm{S})=\frac{1}{36}
\end{aligned}
$
P (Sum of the numbers is divisible by 3 or 4 )
$
\begin{aligned}
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
= & \frac{12}{36}+\frac{9}{36}-\frac{1}{36}=\frac{20}{36}=\frac{5}{9}
\end{aligned}
$
ii) Let event A: Sum of the numbers is divisible by 3
$
\therefore \mathrm{P}(\mathrm{A})=\frac{12}{36}
$
Let event Y: Sum of the numbers is divisible by 5 .
$\therefore$ possible sums are 5,10
$
\begin{aligned}
\therefore & Y=\{(1,4),(2,3),(3,2),(4,1),(4,6), \\
& (5,5),(6,4)\} \\
\therefore & n(Y)=7 \therefore P(Y)=\frac{n(Y)}{n(S)}=\frac{7}{36}
\end{aligned}
$
$\therefore$ Event AnY: sum is divisible by 3 and 5
$
\therefore \mathrm{A} \cap \mathrm{Y}=\phi
$
[X and Y are mutually exclusive events]
$
\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{Y})=\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{Y})}{\mathrm{n}(\mathrm{S})}=0
$
$\therefore$ required probability $=\mathrm{P}($ Sum of the numbers is neither divisible by 3 nor 5 )
$
\begin{aligned}
P\left(A^{\prime} \cap Y^{\prime}\right) & =P(A \cup Y)^{\prime}[\text { De'Morgan's law] } \\
& =1-P(X \cup Y) \quad[\text { Property 1] } \\
& =1-[P(A)+P(Y)-P(A \cap Y)] \\
& =1-\frac{19}{36}=\frac{17}{36}
\end{aligned}
$

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Question 32 Marks

The letters of the word STORY be arranged randomly. Find the probability that
a) T and Y are together.
b) arrangment begins with T and end with Y.

Answer

The word STORY consists of 5 different letters, which can be arranged among themselves in 5 ! ways.
$
\therefore n(\mathrm{~S})=5 !=120
$
a) Event $A$ : $T$ and $Y$ are a together.
Let us consider $\mathrm{T}$ and $\mathrm{Y}$ as a single letter say $\mathrm{X}$. Therefore, now we have four different
letters $\mathrm{X}, \mathrm{S}, \mathrm{O}$ and $\mathrm{R}$ which can be arranged among themselves in $4 !=24$ different ways. After this is done, two letters $\mathrm{T}$ and $\mathrm{Y}$ can be arranged among themselves in $2 !=2$ ways. Therefore, by fundamental principle, total number of arrangements in which $\mathrm{T}$ and $\mathrm{Y}$ are always together is $24 \times 2=48$.
$\therefore$ required probability $\mathrm{P}(\mathrm{A})=\frac{48}{120}=\frac{2}{5}$
b) Event B: An arrangement begins with $\mathrm{T}$ and ends with $\mathrm{Y}$.
Remaining 3 letters in the middle can be arranged in $3 !=6$ different ways.
$\therefore$ required probability $\mathrm{P}(\mathrm{B})=\frac{6}{120}=\frac{1}{120}$

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Question 42 Marks

If A∪B∪C = S (the sample space) and A, B and C are mutually exclusive events, can the
following represent probability assignment?
i) P(A) = 0.2, P(B) = 0.7, P(C) = 0.1
ii) P(A) = 0.4, P(B) = 0.6, P(C) = 0.2

Answer

i) Since P(S) = P(A∪B∪C)
= P(A) + P(B) + P(C) [Property 10]
= 0.2 + 0.7 + 0.1 = 1
and 0≤ P(A), P(B), P(C) ≤ 1
∴ The given values can represent the
probability assignment.
ii) Since
P(S) = P(A∪B∪C)
= P(A) + P(B) + P(C) [Property 10]
= 0.4 + 0.6 + 0.2 = 1.2 > 1
∴ P(A), P(B) and P(C) cannot represent
probability assignment.

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Maths Complete the following activities and rewrite it : (2M)

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