Question

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 1

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)	\(\frac{25}{36}\)	\(\frac{11}{36}\)