Question

A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.

Answer 1

It is given that "success" denotes the event of getting the numbers 1, 3 or 5. Then,

\[P\left( \text{ success } \right) = \frac{1}{2}\]

Also, "failure" denotes the event of getting the numbers 2, 4 or 6. Then,

\[P\left( \text{ failure } \right) = \frac{1}{2}\]

Let X denote the event of getting success.Then, X can take the values 0, 1 and 2.
Now,

\[P\left( X = 0 \right) = P\left( \text{ no success }  \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\]
\[P\left( X = 1 \right) = P\left( 1 \text{ success } \right) = \left( \frac{1}{2} \times \frac{1}{2} \right) + \left( \frac{1}{2} \times \frac{1}{2} \right) = \frac{1}{2}\]
\[P\left( X = 2 \right) = P\left( 2 \text{ success } \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\]

Thus, the probability distribution of X is given by

X	P(X)
0	\[\frac{1}{4}\]
1	\[\frac{1}{2}\]
2	\[\frac{1}{4}\]

Computation of mean and variance

xi	pi	pixi	pixi2
0	\[\frac{1}{4}\]	0	0
1	\[\frac{1}{2}\]	\[\frac{1}{2}\]	\[\frac{1}{2}\]
2	\[\frac{1}{4}\]	\[\frac{1}{2}\]	1
		\[\sum\nolimits_{}^{}\] pixi = 1	\[\sum\nolimits_{}^{}\] pixi2 = \[\frac{3}{2}\]

 

\[\text{ Mean}  = \sum p_i x_i = 1\]
\[\text{ Variance  } = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean}  \right)^2 = \frac{3}{2} - 1 = \frac{1}{2}\]