GATE CS 2018 QGA-10 — General Aptitude

This problem can be modeled using the Binomial Distribution.

Let $G$ be the event of rolling a green face, and $R$ be the event of rolling a red face. The probabilities of these events on a single roll of the unbiased 6-sided die are:

• Probability of Green, $p = P(G) = \frac{4}{6} = \frac{2}{3}$
• Probability of Red, $q = P(R) = \frac{2}{6} = \frac{1}{3}$

The die is rolled $n = 7$ times. Let $X$ be the random variable representing the number of green faces obtained. $X$ follows a binomial distribution, $X \sim \text{Binomial}\left(7, \frac{2}{3}\right)$.

The probability of getting exactly $k$ green faces (and $7-k$ red faces) is given by:

We can find the most likely outcome (the mode of the distribution) by calculating $P(X = k)$ for the values of $k$ given in the options:

1. Three green and four red faces ($k = 3$): $$P(X = 3) = \binom{7}{3} \frac{2^3}{3^7} = 35 \times \frac{8}{3^7} = \frac{280}{3^7}$$
2. Four green and three red faces ($k = 4$): $$P(X = 4) = \binom{7}{4} \frac{2^4}{3^7} = 35 \times \frac{16}{3^7} = \frac{560}{3^7}$$
3. Five green and two red faces ($k = 5$): $$P(X = 5) = \binom{7}{5} \frac{2^5}{3^7} = 21 \times \frac{32}{3^7} = \frac{672}{3^7}$$
4. Six green and one red face ($k = 6$): $$P(X = 6) = \binom{7}{6} \frac{2^6}{3^7} = 7 \times \frac{64}{3^7} = \frac{448}{3^7}$$

Comparing the numerators, we see that the probability is maximized when $k = 5$ (numerator is $672$).

Alternative Method (Using the Mode of a Binomial Distribution):
The mode of a binomial distribution $B(n, p)$ is given by $\lfloor (n+1)p \rfloor$ if $(n+1)p$ is not an integer.
Here, $(n+1)p = (7+1) \times \frac{2}{3} = \frac{16}{3} \approx 5.33$.
Therefore, the mode is $\lfloor 5.33 \rfloor = 5$ green faces (and consequently $7 - 5 = 2$ red faces).