Q. 45.0( 1 Vote )

A die is tossed twice. ‘Getting an odd number on a toss’ is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.

Answer :

Given : A die is tossed twice and ‘Getting an odd number on a toss’ is considered a success.


To find : probability distribution of the number of successes and mean (𝓊) and variance (σ2)


Formula used :



Mean = E(X) =


Variance = E(X2) -


Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)


When a die is tossed twice,


Total possible outcomes =


{(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)


(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)


(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)


(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)


(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)


(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}


‘Getting an odd number on a toss’ is considered a success.


P(0) = = (zero odd numbers = 9 )


P(1) = = (one odd number = 18 )


P(2) = = (two odd numbers = 9 )


The probability distribution table is as follows,



Mean = E(X) = 0() + 1() +2() = 0 + + = = 1


Mean = E(X) = 1


= = 1


E(X2) = = P(x1) + P(x2) + P(x3)


E(X2) = () + () + () = 0 + + = = = 1.5


E(X2) = 1.5


Variance = E(X2) - = 1.5 – 1 = 0.5


Variance = E(X2) - = 0.5


The probability distribution table is as follows,



Mean = 1


Variance = 0.5


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