# If a and b are mutually exclusive events with p(a) = 0.3 and p(b) = 0.5, then p(a ∩ b) =

1. An experiment yields four results, each with $P ( E_1 ) = 0.2$, $P ( E_2 ) = 0.3$ and $P ( E_3 ) = 0.4$. What is the probability of $E_4$?
2. An experiment yields four results, each with $P ( E_1 ) = 0.2$, $P ( E_2 ) = 0.2$ and $P ( E_3 ) = 0.4$. What is the probability of $E_4$?

The main objective of this question is to find the probability of an outcome when two events are mutually exclusive.

This question uses the concept of mutually exclusive events. When two occurrences do not occur simultaneously, such as when a die is thrown or when we flip a coin, they are mutually exclusive. The likelihood that it will land on its head or its tail is entirely independent of one another. These two things cannot happen at the same time; either the head or the tail will come first. Events of this nature are referred to be mutually exclusive events.

1) In this question, we have to find the probability of an event when the two events are mutually exclusive.

We know that when events are mutually exclusive:

$P(A \cap B) \space = \space 0$

And:

$= \space P ( A u B) = \space P ( A ) \space + \space P (B )- P ( A n B )$

By putting values, we get:

$= \space 0.3 \space + \space 0.5 \space – \space 0 \space = \space 0.8$

2) In this question, we have to find the probability of an event which is $E_4$.

So:

We know that sum of probability is equal to $1$.

$P (E4) \space = \space 1 \space – \space 0.2 \space – \space 0.3 \space – \space 0.4 \space = \space 0.1$

3) In this question, we have to find the probability of an event which is E_4.

So:

We know that sum of probability is equal to $1$.

$P (E4) \space = \space 1 \space – \space 0.2 \space – \space 0.2 \space – \space 0.4 \space = \space 0.2$

1. The probability of $a \cap b$ is $0.8$.
2. The probability of event which is $E_4$ is $0.1$.
3. The probability of event which is  $E_4 is$  0.2 $. ## Example An experiment yields four results, each with$ P ( E_1 ) = 0.2  $,$ P ( E_2 ) = 0.2 $and$ P ( E_3 ) = 0.2 $. What is the probability of$E_4 $? Another experiment also yields four results, each with$ P ( E_1 ) = 0.1  $,$ P ( E_2 ) = 0.1 $and$ P ( E_3 ) = 0.1 $. What is the probability of$E_4 $? In this question, we have to find the probability of an event which is$ E_4 $. So: We know that sum of probability is equal to$1 $. $P (E4) \space = \space 1 \space – \space 0.2 \space – \space 0.2 \space – \space 0.2 \space = \space 0.4$ Now for the second experiment we have to find the probability of an event which is$E_4 $. So: We know that sum of probability is equal to$1\$.
$P (E4) \space = \space 1 \space – \space 0.1 \space – \space 0.1 \space – \space 0.1 \space = \space 0.7$