Question

Let $E1,E2,E3$ be three arbitrary events of a sample space $S$ , consider the following statements, which of the following statements are correct.
1. P (only one of them occurs) $= P(E1'E2E3 + E1E2'E3 + E1E2E3')$
2. P (none of them occurs) $P(E1' + E2' + E3')$
3. P (at least one of them occurs) $= P(E1 + E2 + E3)$
4. P (all three of them occurs) $= P(E1 + E2 + E3)$

Answer 1

In this question, we are given three events $E1,E2,E3$ and the sample space $S$ and we have to find the probability according to the options given in the options.
One must remember that “only one of them occurs” means exactly one of them will occur, and “at least one of them occurs” means one or more than one can also occur.

Formula to be used:
Simple formula that we know of probability is given by $\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$ .
$P(E1 \cup E2 \cup E3) = P(E1) + P(E2) + P(E3) - P(E1 \cap E2) - P(E2 \cap E3) - P(E1 \cap E3) + P(E1 \cap E2 \cap E3)$
Probability of an event to occur is given by $P(E)$ and probability for an event not to occur is given by $P(E') = 1 - P(E)$ .

Complete step-by-step answer:
Given sample space $S$ and the three events $E1,E2,E3$ .
We know that probability is referred to as the ‘chance of an event to happen’.
We can write $P(E1 + E2 + E3) = P(E1) + P(E2) + P(E3)$ , only if $E1,E2,E3$ are mutually exclusive events.
Now, first of all probability that at least one of them will occur is given by
$P(E1 \cup E2 \cup E3)$ , which can also be written as $P(E1 + E2 + E3)$ .
So, option $(3)$ is correct.

Now, Probability that all of them will occur is given by $= P(E1 \cap E2 \cap E3)$ .
So, option $(4)$ is not correct.
Now, probability of none of them occurred is same as $1$ $-$ at least one of them occurred, i.e., P (none of them occurred) $=$ $1$ $-$ P (at least one of them occurred) which is $P(E1' \cap E2' \cap E3') = 1 - P(E1 + E2 + E3)$ .
So, option $(2)$ is also incorrect.

Now, probability of only one of the three events will occur which also means the other two will not occur is given by $P(E1 \cap E2' \cap E3') + P(E1' \cap E2 \cap E3') + P(E1' \cap E2' \cap E3)$
So, option $(1)$ is also incorrect.
Finally, the probability of at least one of them occurring is given by $= P(E1 + E2 + E3)$ .

So, the correct answer is “Option (3)”.

Note: When we have to take elements from $A$ or $B$ , then we take probability of union of the two sets, whereas, when we have to take elements from $A$ and $B$ , then, we take probability of intersection of the two sets.
The point to remember is that ‘none of them occurred’ is not a compliment of ‘all of them occurred’ but it is opposite of ‘at least one of them occurred’.
Another point is ‘at least one of them occurred’ implies one has to occur with more than one, even all of them can also occur but ‘none of them’ will not be the case, whereas ‘only one of them occurred’ implies exactly one of them has to occur and not more than that.