GATE CSE 1999 | Question: 2.1

Question

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?

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

• B. Events $E_1$ and $E_2$ are independent

• C. Events $E_1$ and $E_2$ are not independent

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

Comments

which canvas is this

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@goku4199 in conditional probability intersection is given which is 1/5, so conditional probability answer is 3/5. 

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Answer 1

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

Comments

You should also add why $E_1$ and $E_2$ are not independent.

Let's assume $E_1$ and $E_2$ are independent.

So,

$P(E_1 \wedge E_2) = P(E_1) \cdot P(E_2)  \implies \left( \frac{1}{5} \right) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{3} \right) \implies   \left( \frac{1}{5} \right) = \left( \frac{1}{6} \right)$

which is a contradiction.

Hence, $E_1$ and $E_2$ are not independent.

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Isn't that already there in answer?

Answer 2

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.

Answer 3

Ans