Answer :

Given : A die is tossed twice and ‘Getting a number greater than 4 ’ 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) -


When a die is tossed 4 times,


Total possible outcomes = 62 = 36


Getting a doublet is considered as a success


The possible doublets are (1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)


Let p be the probability of success,


p = =


q = 1 – p = 1 - =


q =


since the die is thrown 4 times, n = 4


x can take the values of 1,2,3,4


P(x) = nCx


P(0) = 4C0 =


P(1) = 4C1 = =


P(2) = 4C2 = =


P(3) = 4C3 = =


P(4) = 4C4 =


The probability distribution table is as follows,



Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5)


Mean = E(X) = 0() + 1() + 2() + 3() + 4()


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


Mean = E(X) =


= =


E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5)


E(X2) = () + () + () + () + ()


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


E(X2) = 1


Variance = E(X2) - = 1 – =


Variance = E(X2) - =


The probability distribution table is as follows,



Mean =


Variance =


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