Prove that if $E$ and $F$ are independent events, then so are the events $\mathrm{E}$ and $\mathrm{F}^{\prime}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

since $\mathrm{E}$ and $\mathrm{F}$ are independent, we have

$\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})$       ......... $(1)$

From the venn diagram in Fig it is clear that $E \cap \mathrm{F}$ and $\mathrm{E} \cap \mathrm{F}^{\prime}$ are mutually exclusive events and also $\mathrm{E}=(\mathrm{E} \cap \mathrm{F}) \cup\left(\mathrm{E} \cap \mathrm{F}^{\prime}\right)$

Therefore        $\quad P(E)=P(E \cap F)+P\left(E \cap F^{\prime}\right)$

or                   $P\left(E \cap F^{\prime}\right)=P(E)-P(E \cap F)$

                    $=\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})$    (by $(1))$

                   $=\mathrm{P}(\mathrm{E})(1-\mathrm{P}(\mathrm{F}))$

                   $=\mathrm{P}(\mathrm{E})$ . $\mathrm{P}\left(\mathrm{F}^{\prime}\right)$

Hence, $\mathrm{E}$ and $\mathrm{F}^{\prime}$ are independent

863-s41

Similar Questions

If an integer is chosen at random from first $100$ positive integers, then the probability that the chosen number is a multiple of $4$ or $6$, is

$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively

The probabilities of three mutually exclusive events are $\frac{2}{3} ,  \frac{1}{4}$ and $\frac{1}{6}$. The statement is

If $A$ and $B$ are two events of a random experiment, $P\,(A) = 0.25$, $P\,(B) = 0.5$ and $P\,(A \cap B) = 0.15,$ then $P\,(A \cap \bar B) = $

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\frac{1}{3}$, then the probability that the experiment stops with head is.

  • [IIT 2023]