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# Find the Probability Distribution of the Number of Successes in Two Tosses of a Die, Where a Success is Defined as - Mathematics and Statistics

#### Question

Find the probability distribution of the number of successes 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

#### Solution

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.

Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6xx4/6 = 16/36 = 4/9

P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)

= 4/6xx2/6+4/6xx2/6=8/36 + 8/36 = 16/36 =4/9

P (X = 2) = P (number greater than 4 on both the tosses)

= 2/6xx2/6= 4/36 =1/9

Thus, the probability distribution is as follows.

 X 0 1 2 P(X) 4/9 4/9 1/9

(ii) Here, success means six appears on at least one die.

P (Y = 0 ) = P (six appears on  none of the dice) = 5/6 xx 5/6 = 25/36

P(Y = 1) = P (six appears on  none of the dice x six appears on at least one of the dice ) + P (six appears on  none of the dice x six appears on at least one of the dice)

= 1/6 xx 5/6 + 1/6 xx 5/6 = 5/36 + 5/36 = 10/36

P (Y = 2) = P (six appears on at least one of the dice) = 1/6 xx 1/6 =1/36

Thus, the required probability distribution is as follows

 Y 0 1 2 P(Y) 25/36 10/36 1/36
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