Question

In an assembly of $4$persons the probability that at least 2 of them have the same birthday, is
1. $0.293$
2. $0.24$
3. $0.0001$
4. $0.016$

Answer 1

Total number of days in a year is $365$. Now calculate the probability of $2$ persons having their birthday on the same day using the concept of combination and multiplication theorem. Similarly for $3$ persons and for $4$ persons. Then add all the probability to get the required or total probability.

Complete step by step answer:
We all know that the probability of any event to occur is always greater than or equal to $0$ or less than or equal to $1$. In the given question we have asked about the probability that at least 2 of them have the same birthday when $4$ persons are there in the assembly.
The word at least means that not less than that implies always equal to or greater than the given value. So according to the question the total probability is calculated as
$\text{Total Probability}=P({{E}_{1}})+P({{E}_{2}})+P({{E}_{3}})$
Where ,
${{E}_{1}}=2$persons having a birthday on the same day.
${{E}_{2}}=3$ persons having a birthday on the same day.
${{E}_{3}}=4$ persons having a birthday on the same day.
So let us calculate all the probability one by one. In general there are a total of$365$ days. So the total number of outcomes is $365$.
If we are calculating for $2$persons having the birthday on the same day then out of $4$ persons we have to select $2$ of them. And using the multiplication theorem to get the probabilities.
Probability of having the birthday on same day $=\dfrac{1}{365}$
Probability of having the birthday on different day $=\dfrac{364}{365}$
$P({{E}_{1}}){{=}^{4}}{{C}_{2}}(\dfrac{1}{365})\times {{(\dfrac{364}{365})}^{2}}$
Similarly we can say that,
$P({{E}_{2}}){{=}^{4}}{{C}_{3}}{{(\dfrac{1}{365})}^{2}}\times (\dfrac{364}{365})$
$P({{E}_{3}}){{=}^{4}}{{C}_{4}}{{(\dfrac{1}{365})}^{3}}$
Now add all the probabilities to get the Total Probability or we can say required probability
$\text{Total Probability}=P({{E}_{1}})+P({{E}_{2}})+P({{E}_{3}})$
$\text{Total Probability}{{=}^{4}}{{C}_{2}}(\dfrac{1}{365})\times {{(\dfrac{364}{365})}^{2}}{{+}^{4}}{{C}_{3}}{{(\dfrac{1}{365})}^{2}}\times (\dfrac{364}{365}){{+}^{4}}{{C}_{4}}{{(\dfrac{1}{365})}^{3}}$
To solve the combination, we have to use the formula
$^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$
Where,
$n$is the total number of elements.
$r$is the number of elements we want to select.
Simplifying the above expression, we get
$\Rightarrow \text{Total Probability}=\dfrac{6\times {{(364)}^{2}}+4\times 364+1}{{{(365)}^{3}}}$
$\Rightarrow \text{Total Probability}=\dfrac{796433}{48627125}$
$\Rightarrow \text{Total Probability}=0.016$
So we can say that in an assembly of $4$persons the probability that at least 2 of them have the same birthday is $0.016$

So, the correct answer is “Option 4”.

Note: If the two events are independent events i.e. the events in which the occurrence of one event does not impact the occurrence of the second event, then the probability of both the events occurring is equal to the product of their individual probabilities; this theorem is known as the multiplication theorem.