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

Find the probability distribution of the number of successes in two to

1.	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
Answer» Here, the total number of outcomes,`n(S) = 6^2 = 36` (i) Number of outcomes when both tosses of die do not have a number greater than 4,(X=0)`=44=16` Number of outcomes when exactly one toss of die has a number greater than 4,(X=1)`=42+24=16` Number of outcomes when both tosses of die have a number greater than 4(X=2),`=22 = 4` So, `P(X=0) = 16/36 = 4/9` `P(X=1) = 16/36 = 4/9` `P(X=2) = 4/36 = 1/9` (ii) Number of outcomes when six does not appear on both tosses of die,(X=0)`=55=25` Number of outcomes when six appear only on one toss of die,(X=1)`=51+15=10` Number of outcomes when six appear on both tosses of die,4(X=2),`=11 = 4` So, `P(X=0) = 25/36` `P(X=1) = 10/36 = 5/18` `P(X=2) = 1/36 = 1/36`

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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(i) number greater than 4(ii) six appears on at least one die

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