Question

Find the probability distribution of the number of successes in two tosses of a die if success is defined as getting a number greater than 4.

Answer 1

Success is defined as a number greater than 4 appears on at least one die.

Let X denote the number of successes.

∴ Possible values of X and 0, 1, 2.

Let, P(getting a number greater than 4) = p = `(2)/(6) = (1)/(3)`

∴ q = 1 – p = `1 - (1)/(3) = (2)/(3)`

∴ P(X = 0) = No number greater than 4 in two tosses of dice.

∴ P(X = 0) = P {(1, 1) (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

`= 16/36 = 4/9`

∴ P(X = 1) = Only 1 number is greater than 4 in two tosses of o die.

∴ P(X = 1) = P(one success) = P {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}

`= 16/36`

= `(4)/(9)`

∴ P(X = 2) = Two numbers are greater than 4 in two tosses of die.

∴ P(X = 2) = P {(5, 5), (5, 6), (6, 5), (6, 6)}

`= 4/36`

= `1/9`

∴ Probability distribution of X is as follows:

X	0	1	2
P(X = x)	`(4)/(9)`	`(4)/(9)`	`(1)/(9)`