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 (σ^{2})

Formula used :

Mean = E(X) =

Variance = E(X^{2}) -

When a die is tossed 4 times,

Total possible outcomes = 6^{2} = 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) = ^{n}C_{x}

P(0) = ^{4}C_{0} =

P(1) = ^{4}C_{1} = =

P(2) = ^{4}C_{2} = =

P(3) = ^{4}C_{3} = =

P(4) = ^{4}C_{4} =

The probability distribution table is as follows,

Mean = E(X) = = x_{1}P(x_{1}) + x_{2}P(x_{2}) + x_{3}P(x_{3}) + x_{4}P(x_{4}) + x_{5}P(x_{5})

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

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

Mean = E(X) =

= =

E(X^{2}) = = P(x_{1}) + P(x_{2}) + P(x_{3}) + P(x_{4}) + P(x_{5})

E(X^{2}) = () + () + () + () + ()

E(X^{2}) = 0 + + + + = =

E(X^{2}) = 1

Variance = E(X^{2}) - = 1 – =

Variance = E(X^{2}) - =

The probability distribution table is as follows,

Mean =

Variance =

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