Question

A die is tossed twice. ‘Getting a number greater than 4’ 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 1

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 (μ) and variance (σ²)

Formula used :

When a die is tossed twice,

Total possible outcomes =

‘Getting a number greater than 4’ is considered a success.

P(0) = \(\frac{16}{36}=\frac{4}{9}\) (zero numbers greater than 4 = 16)

P(1) = \(\frac{16}{36}=\frac{4}{9}\) (one number greater than 4 = 16)

P(2) = \(\frac{4}{36}=\frac{1}{9}\) (two numbers greater than 4= 4)

The probability distribution table is as follows,

The probability distribution table is as follows,

Mean = \(\frac{2}{3}\)

Mean = \(\frac{4}{9}\)