Question

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

Hint:

For a binomial distribution \( B(n, p) \), the mean is given by \( E(X) = n p \).

Answer 1

To solve the problem, we are given a biased die with numbers from 1 to 6.

1. Given:
- \( P(2) = \frac{3}{10} \)
- The probabilities of the other numbers (1, 3, 4, 5, 6) are equal.
- Total probability must be 1.
Let the probability of each of the other 5 numbers be \( p \). Then:
\[ P(2) + 5p = 1 \Rightarrow \frac{3}{10} + 5p = 1 \Rightarrow 5p = \frac{7}{10} \Rightarrow p = \frac{7}{50} \]

2. Define Random Variable:
Let \( X \) be the number of times the number 2 appears in 2 throws of the die.
This is a binomial distribution with:
- Number of trials \( n = 2 \)
- Probability of success (i.e., getting a 2) \( p = \frac{3}{10} \)

3. Mean of a Binomial Distribution:
The mean is given by:
\[ E(X) = n \cdot p = 2 \cdot \frac{3}{10} = \frac{6}{10} = \frac{3}{5} \]

Final Answer:
The mean number of times 2 appears in two throws is \( \boxed{\frac{3}{5}} \).