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 (σ^{2})

Formula used :

Mean = E(X) =

Variance = E(X^{2}) -

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

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

E(X^{2}) = () + () + () = 0 + + = = = 1.5

E(X^{2}) = 1.5

Variance = E(X^{2}) - = 1.5 – 1 = 0.5

Variance = E(X^{2}) - = 0.5

The probability distribution table is as follows,

Mean = 1

Variance = 0.5

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