TIFR CSE 2020 | Part A | Question: 11

Question

Suppose we toss $m=3$ labelled balls into $n=3$ numbered bins. Let $A$ be the event that the first bin is empty while $B$ be the event that the second bin is empty. $P(A)$ and $P(B)$ denote their respective probabilities. Which of the following is true?

• A. $P(A)>P(B)$
• B. $P(A) = \dfrac{1}{27}$
• C. $P(A)>P(A\mid B)$
• D. $P(A)<P(A\mid B)$
• E. None of the above

Answer 1

Suppose, we have $3$ labelled balls as $\{B_1,B_2,B_3\}$ and $3$ numbered bins as $\{U_1,U_2,U_3\}.$ Now, number of ways to throw these balls into bins is same as total number of functions from set $\{B_1,B_2,B_3\}$ to set $\{U_1,U_2,U_3\}.$ So, sample size = $3^3$

Now, $A$ be the event that first bin is empty and we have to throw balls into bins. So, number of favorable outcomes is total number of functions from set $\{B_1,B_2,B_3\}$ to set $\{U_2,U_3\}$ which is $2^3.$ So, $\mathbb{P}(A) = \frac{2^3}{3^3}.$

Now, $B$ be the event that second bin is empty and we have to throw balls into bins. So, in this case, number of favorable outcomes is total number of functions from set $\{B_1,B_2,B_3\}$ to set $\{U_1,U_3\}$ which is $2^3.$ So, $\mathbb{P}(B) = \frac{2^3}{3^3}.$

Now, $\mathbb{P}(A|B)$ means given second bin is empty, what is the probability that first bin is empty. So, $\mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B )}{\mathbb{P}(B)} = \frac{\frac{1}{3^3}}{\frac{2^3}{3^3}}=\frac{1}{2^3}$

So, $\mathbb{P}(A) \neq \frac{1}{27}$, $\mathbb{P}(A) = \mathbb{P}(B)$ and $\mathbb{P}(A) > \mathbb{P}(A|B)$