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

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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.
Updated On: Jun 8, 2026
  • \( \frac{2}{9} \)
  • \( \frac{2}{3} \)
  • \( \frac{3}{4} \)
  • \( \frac{4}{9} \)
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The Correct Option is D

Solution and Explanation

Concept: Repeated independent trials with fixed success probabilities follow a Binomial Distribution. The variance of a binomial distribution is given by the formula: \[ \text{Variance} = n p q \] where \( n \) is the number of trials, \( p \) is the probability of success, and \( q = 1 - p \) is the probability of failure.

Step 1: Determining single-trial parameters.
A standard fair die has 6 outcomes: \( \{1, 2, 3, 4, 5, 6\} \). Success is defined as rolling a number strictly greater than 4, which includes only \( \{5, 6\} \) (2 outcomes). \[ p = \frac{2}{6} = \frac{1}{3} \] The probability of failure \( q \) is: \[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

Step 2: Calculating the variance for the given number of trials.
The die is thrown twice, so the number of trials is \( n = 2 \). \[ \text{Variance} = n p q = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9} \]
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