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.
Y | 0 | 1 |
P(Y) | \(\frac{25}{36}\) | \(\frac{11}{36}\) |
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