GATE CS 2018 Set 1 — Question 10

This is a problem of binomial distribution.
Let G be the event of getting a green face and R be the event of getting a red face.
The die has 6 faces in total.
Number of green faces = 4
Number of red faces = 2Probability of getting a green face, $p = P(G) = \frac{4}{6} = \frac{2}{3}$.
Probability of getting a red face, $q = P(R) = \frac{2}{6} = \frac{1}{3}$.The die is rolled 7 times, so the number of trials is $n=7$.Let $k$ be the number of green faces obtained in 7 rolls. The number of red faces will be $7-k$. The random variable $k$ follows a binomial distribution with parameters $n=7$ and $p=2/3$.
The probability of getting exactly $k$ green faces is given by the formula:
$P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{7}{k} (\frac{2}{3})^k (\frac{1}{3})^{7-k}$The most likely outcome corresponds to the mode of the binomial distribution. The mode for a binomial distribution $B(n, p)$ is the integer $k$ that satisfies $(n+1)p - 1 \le k \le (n+1)p$.Let's calculate the value of $(n+1)p$:
$(n+1)p = (7+1) \times \frac{2}{3} = 8 \times \frac{2}{3} = \frac{16}{3} = 5.33$So, the mode is $k = \lfloor (n+1)p \rfloor = \lfloor 5.33 \rfloor = 5$.
This means the most likely number of green faces is 5. If there are 5 green faces, the number of red faces must be $7 - 5 = 2$.Therefore, the most likely outcome is 5 green faces and 2 red faces.Alternatively, we can calculate the probability for each option:
(A) 3 green, 4 red ($k=3$): $P(X=3) = \binom{7}{3} (\frac{2}{3})^3 (\frac{1}{3})^4 = 35 \times \frac{8}{2187} = \frac{280}{2187}$
(B) 4 green, 3 red ($k=4$): $P(X=4) = \binom{7}{4} (\frac{2}{3})^4 (\frac{1}{3})^3 = 35 \times \frac{16}{2187} = \frac{560}{2187}$
(C) 5 green, 2 red ($k=5$): $P(X=5) = \binom{7}{5} (\frac{2}{3})^5 (\frac{1}{3})^2 = 21 \times \frac{32}{2187} = \frac{672}{2187}$
(D) 6 green, 1 red ($k=6$): $P(X=6) = \binom{7}{6} (\frac{2}{3})^6 (\frac{1}{3})^1 = 7 \times \frac{64}{2187} = \frac{448}{2187}$Comparing the probabilities:
$rac{672}{2187} > \frac{560}{2187} > \frac{448}{2187} > \frac{280}{2187}$
The highest probability is for 5 green faces and 2 red faces.