CBSE 2025, Set 2

Question

A die with number 1 to 6 is biased such that $P(2)=\frac{3}{10}$ and probability of other numbers is equal. Find the mean of the number of times number 2 appears on the dice, if the dice is thrown twice.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1: Determine the individual probabilities.** Given that the probability of getting a '2' is $P(2) = \frac{3}{10}$. Let the event of getting a '2' be considered a "success" ($p$). $p = P(2) = \frac{3}{10}$ The probability of a "failure" (not getting a '2') is $q = 1 - p$. $q = 1 - \frac{3}{10} = \frac{7}{10}$ **Step 2: Set up the random variable.** The dice is thrown twice, so $n = 2$. Let $X$ be the random variable denoting the number of times '2' appears. $X$ can take values $0, 1, 2$. This follows a Binomial Distribution: $X \sim B(n=2, p=0.3)$. **Step 3: Calculate the mean.** The mean (Expected Value) of a Binomial distribution is given directly by the formula $\mu = n \cdot p$. $$\mu = 2 	imes \frac{3}{10} = \frac{6}{10} = 0.6$$ *(Alternative Method via Probability Distribution):* $P(X=0) = \binom{2}{0} p^0 q^2 = 1 \cdot (\frac{7}{10})^2 = \frac{49}{100}$ $P(X=1) = \binom{2}{1} p^1 q^1 = 2 \cdot (\frac{3}{10}) \cdot (\frac{7}{10}) = \frac{42}{100}$ $P(X=2) = \binom{2}{2} p^2 q^0 = 1 \cdot (\frac{3}{10})^2 = \frac{9}{100}$ $	ext{Mean } E(X) = \sum X \cdot P(X)$ $= 0\left(\frac{49}{100}ight) + 1\left(\frac{42}{100}ight) + 2\left(\frac{9}{100}ight)$ $= 0 + \frac{42}{100} + \frac{18}{100} = \frac{60}{100} = 0.6$ The mean is **$0.6$**.

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