Find the probability distribution of the number of success in two tosses of a die where a success is defined as: (i)number greater than 4 (ii)six appears on at least one die.
Answer When a die is tossed two times, we obtain (6 × 6) = 36 number of observations. Let X be the random variable, which represents the number of successes. i. Here, success refers to the number greater than 4. P (X = 0) = P (number less than or equal to 4 on both the tosses) =\(\frac{2}{6}\times\frac{2}{6}=\frac{1}{9}\) Thus, the probability distribution is as follows.
X
0
1
2
P(X)
\(\frac{4}{9}\)
\(\frac{4}{9}\)
\(\frac{1}{9}\)
(ii) Here, success means six appears on at least one die. P (Y = 0) = P (six does not appear on any of the dice)=\(\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}\) P (Y = 1) = P (six appears on at least one of the dice) =\(\frac{11}{36}\) Thus, the required probability distribution is as follows.
