2 Marks Questions

Question 12 Marks

A die is thrown. If $E$ is the event 'the number appearing is a multiple of $3^{\prime}$ and $F$ be the event 'the number appearing is even' then find whether $E$ and $F$ are independent?

Answer

We know that the sample space of given experiment is
$
S=\{1,2,3,4,5,6\}
$
Now $\quad E=\{3,6\}, F=\{2,4,6\}$ and $E \cap F=\{6\}$
then $\quad P ( E )=\frac{2}{6}=\frac{1}{3}, P ( F )=\frac{3}{6}=\frac{1}{2}$
and $P ( E \cap F )=\frac{1}{6}$
Clearly $\quad P ( E \cap F )= P ( E ) \cdot P ( F )$
Hence E and F are independent events.

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Question 22 Marks

A fair die has been tossed. Find $P(E / F)$ and $P(F / E)$ for the events $E =\{1,3,5\}, F =\{2,3\}$ and $G =\{2,3,4,5\}$.

Answer

A die can show $1,2,3,4,5$ or 6 . Therefore there are 6 results in sample space.
$
\therefore \quad \begin{aligned}
n(S) & =6 \\
E & =\{1,3,5\}, F\{2,3\}, G\{2,3,4,5\}
\end{aligned}
$
$\begin{aligned} E \cap F & =\{3\} \Rightarrow n( E \cap F )=1 \\ \therefore \quad P ( E \cap F ) & =\frac{1}{6}, P ( E )=\frac{3}{6}, P ( F )=\frac{2}{6} \\ \text { Now } \quad P ( E / F ) & =\frac{ P ( E \cap F )}{ P ( F )}=\frac{1}{6} \div \frac{2}{6}=\frac{1}{2} \\ P \left(\frac{ F }{ E }\right) & =\frac{ P ( E \cap F )}{ P ( E )}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}\end{aligned}$

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Question 32 Marks

In a pack of 52 cards, one by one, two cards are drawn randomly without replacement. Find the probability of both card will be red in colour.

Answer

ATQ, $\quad$ total cards in a pack $=52$
Number of red colour cards $=26$
robability of drawing one red card is $P ( A )=\frac{26}{52}=\frac{1}{2}$
After one card is drawn, there is 51 cards in which 25 are red.
$\therefore$ Probability of drawing second red card $=\frac{25}{51}$ or without replacement probability of getting 2 red cards
$
=\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}
$

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Question 42 Marks

If A and B are independent events, where $P ( A )=\frac{1}{2}$, $P ( A \cup B )=\frac{3}{5}$ and $P ( B )=x$, then find the value of $x$.

Answer

Because A and B are independent events.
hence $P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )$
or $\quad P ( A \cup B )= P ( A )+ P ( B )- P ( A ) \cdot P ( B )$
or $\quad \frac{3}{5}=\frac{1}{2}+x-\frac{1}{2} \times x$
or $\quad \frac{3}{5}=\frac{1}{2}+\frac{x}{2} \Rightarrow \frac{x}{2}=\frac{3}{5}-\frac{1}{2}$
$\begin{array}{rlrl} & \frac{x}{2} & =\frac{6-5}{10}=\frac{1}{10} \\ \therefore & & x =\frac{2}{10}=\frac{1}{5}\end{array}$

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