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A die with number 1 to 6 is biased such that probability of $P(2)=\frac{3}{10}$ 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.
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Q: A die with number 1 to 6 is biased such that probability of $P(2)=\frac{3}{10}$ 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.

Question Analysis & Solution

Detailed Solution

Step 1: Calculate the probability of numbers other than 2.

Let the probability of each number other than 2 be $p$. Since the sum of all probabilities must be 1, we have: $P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$ $p + \frac{3}{10} + p + p + p + p = 1$ $5p + \frac{3}{10} = 1$ $5p = 1 - \frac{3}{10} = \frac{7}{10}$ $p = \frac{7}{50}$ So, $P(1) = P(3) = P(4) = P(5) = P(6) = \frac{7}{50}$

Step 2: Define the random variable and its distribution.

Let $X$ be the random variable representing the number of times the number 2 appears when the die is thrown twice. $X$ can take values 0, 1, or 2. This is a binomial distribution with $n=2$ trials and probability of success (getting a 2) $P(2) = \frac{3}{10}$.

Step 3: Calculate the probabilities for X = 0, 1, and 2.

The probability mass function for a binomial distribution is given by: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$ In our case, $n=2$ and $p = \frac{3}{10}$. $P(X=0) = \binom{2}{0} (\frac{3}{10})^0 (1-\frac{3}{10})^{2-0} = 1 \cdot 1 \cdot (\frac{7}{10})^2 = \frac{49}{100}$ $P(X=1) = \binom{2}{1} (\frac{3}{10})^1 (\frac{7}{10})^{2-1} = 2 \cdot \frac{3}{10} \cdot \frac{7}{10} = \frac{42}{100}$ $P(X=2) = \binom{2}{2} (\frac{3}{10})^2 (\frac{7}{10})^{2-2} = 1 \cdot (\frac{9}{100}) \cdot 1 = \frac{9}{100}$

Step 4: Calculate the mean of the distribution.

The mean of a binomial distribution is given by $\mu = np$. In our case, $n=2$ and $p = \frac{3}{10}$. Therefore, the mean is: $\mu = 2 \cdot \frac{3}{10} = \frac{6}{10} = \frac{3}{5} = 0.6$

Final Answer: 0.6

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