Question

A dice is thrown twice. If getting a number greater than four is considered a success, the variance of the probability distribution of the number of successes is

• A. \( \frac{2}{9} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{4}{9} \)

Hint:

For any binomial distribution problem, always identify the three core parameters first: \( n \) (total trials), \( p \) (success probability), and \( q \) (failure probability). Once you have these, finding statistical values like the mean (\( np \)) or variance (\( npq \)) becomes incredibly simple.

Answer 1

The Correct Option is D

Step 1: Identify the distribution.
Repeated independent trials with a fixed success chance form a binomial distribution, whose variance is $npq$.
Step 2: Find the success chance.
A die has 6 faces; success means greater than 4, which is $\{5,6\}$, so $p = \tfrac{2}{6} = \tfrac{1}{3}$.
Step 3: Find the failure chance.
$q = 1 - p = \tfrac{2}{3}$.
Step 4: Set the number of trials.
The die is thrown twice, so $n = 2$.
Step 5: Apply the variance formula.
$\text{Variance} = npq = 2\times\tfrac{1}{3}\times\tfrac{2}{3}$.
Step 6: Compute the value.
\[ \text{Variance} = \frac{4}{9} \] \[ \boxed{\tfrac{4}{9}} \]