Question

In an assembly of $4$ persons the probability that at least $2$ of them have the same birthday, is:

• A. $0.293$
• B. $0.24$
• C. $0.0001$
• D. $0.016$

Answer 1

The correct option is D

$0.016$

Explanation for the correct option:

Finding the probability:

No. of days in a non leap year is $365$. So, if out of $n$, two persons have the same birthday then the remaining $n-2$ persons will have their birthday on any of the remaining $364$ days.

So, the required probability will be $$\begin{gathered} =P\left(2\mathrm{people}\mathrm{have}\mathrm{birthday}\mathrm{on}\mathrm{same}\mathrm{day}\right)+P\left(3\mathrm{people}\mathrm{have}\mathrm{birthday}\mathrm{on}\mathrm{same}\mathrm{day}\right)+P\left(4\mathrm{people}\mathrm{have}\mathrm{birthday}\mathrm{on}\mathrm{same}\mathrm{day}\right) \\ ={}^{4}C_2\times {\left(\frac{1}{365}\right)}^1\times {\left(\frac{364}{365}\right)}^2+{}^{4}C_3\times {\left(\frac{1}{365}\right)}^2\times {\left(\frac{364}{365}\right)}^1+{}^{4}C_4\times {\left(\frac{1}{365}\right)}^3\times {\left(\frac{364}{365}\right)}^0 \\ =\frac{4\times 3\times 2!}{2\times 2!}\times \frac{{364}^2}{{\left(365\right)}^3}+\frac{4\times 3!}{3!}\times \frac{364}{{\left(365\right)}^3}+1\times \frac{1}{{\left(365\right)}^3}\left[\because {}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =\frac{6\times {364}^2+4\times 364+1}{{\left(365\right)}^3} \\ =\frac{7,96,433}{48,627,125} \\ \approx 0.016 \end{gathered}$$

Hence, option D is correct.