Question

A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is

• A. $\frac { 2 } { 9 }$
• B. $\frac { 4 } { 9 }$
• C. $\frac { 1 } { 3 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 4 } { 9 }$

Answer 2

Obviously, $p = \frac { 2 } { 6 } = \frac { 1 } { 3 } \Rightarrow q = 1 - \frac { 1 } { 3 } = \frac { 2 } { 3 }$

Also $n = 2$ Therefore, variance $= n p q = 2 \times \frac { 1 } { 3 } \times \frac { 2 } { 3 } = \frac { 4 } { 9 }$.