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Consider two events $E_1$ and $E_2$ such that probability of $E_1$, $P_r[E_1]=\frac{1}{2}$, probability of $E_2$, $P_r[E_{2}]=\frac{1}{3}$, and probability of $E_1$, and $E_2$, $P_r[E_1 \: and \: E_2] = \frac{1}{5}$. Which of the following statements is/are true?

1. $P_r[E_1\: \text{or} \:E_2] \text{ is } \frac{2}{3}$

2. Events $E_1$ and $E_2$ are independent

3. Events $E_1$ and $E_2$ are not independent

4. $P_r \left[{E_1}\mid{E_2} \right] = \frac{4}{5}$

edited | 693 views

+1 vote
For $A$:

$P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2)$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}$

$=\frac{19}{30}$ $\neq \frac{2}{3}$    $\therefore$ $A$ is not True

For$B$:

If $E_1$ and $E_2$ are independent then

$P(E_1\cap E_2)=P(E_1)P(E_2)$

$=\frac{1}{2}\times\frac{1}{3}$

$=\frac{1}{6}$  $\neq \frac{1}{5}$      $\therefore$ $B$ is not True

For $D$:

$P\left ( \frac{E_1}{E_2} \right )=\frac{P(E_1\cap E_2)}{P(E_2)}$

$=\frac{\frac{1}{5}}{\frac{1}{3}}=\frac{3}{5} \neq \frac{4}{5}$  $\therefore$ $D$ is not True

So, the answer is $C$, $E_1$ and $E_2$ are not independent
answered ago by Active (2.9k points)
edited ago by
Answer - $C$

If events $E_1$ and $E_2$ are independent then $P[E_1$ and $E_2]$ = $P[E_1]\times P[E_2]$ which is not the case here.
edited

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