TIFR CSE 2020 | Part A | Question: 7

Question

A lottery chooses four random winners. What is the probability that at least three of them are born on the same day of the week? Assume that the pool of candidates is so large that each winner is equally likely to be born on any of the seven days of the week independent of the other winners.

• A. $\dfrac{17}{2401} \\$
• B. $\dfrac{48}{2401} \\$
• C. $\dfrac{105}{2401} \\$
• D. $\dfrac{175}{2401} \\$
• E. $\dfrac{294}{2401}$

Answer 1

$\\P(\text{exactly }3) = \binom{4}{3}*1*\frac{1}{7}*\frac{1}{7}*\frac{6}{7} = \frac{168}{2401}$

$P(\text{exactly } 4) = \binom{4}{4}*1*\frac{1}{7}*\frac{1}{7}*\frac{1}{7} = \frac{7}{2401}$

$P(\text{atleast } 3) = P(\text{exactly }3) + P(\text{exactly } 4) \\P(\text{atleast }3) = \frac{168}{2401} + \frac{7}{2401} = \frac{175}{2401}$

P(exactly 3) means Probability of exactly 3 people out of 4 having their birthdays on the same day of the week. And so on.

Comments

Multiply (7 C 1 ) in choosing a day then answer is right..

Answer 2

Using Poisson Approximation :

Here, $\lambda =\binom{4}{3}* \frac{1}{7*7}$

So, ${\mathbb{P}}$$($at least 3 of them are born on the same day of the week$)$ $\approx 1-e^{-\lambda} = 1-e^{\binom{4}{3}* \frac{1}{7*7}} = 0.0787$ which is close to option $(d)\; 0.072$. Other Options : $(a) = 0.007 , (b) =0.019, (c) = 0.043, (e) = 0.122$