Solution

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To find the expected number of throws until we get a 3 when throwing a fair three-faced die, we can use the concept of expected value.
Let \( X \) be the random variable representing the number of throws until we get a 3. We need to find \( E(X) \), the expected value of \( X \).
Let's break down the possible scenarios:

1. If the first throw results in a 3, then \( X = 1 \).
2. If the first throw results in a number other than 3:
   • The probability of this happening is \( \frac{2}{3} \) since there are 2 out of 3 faces that are not 3.
   • In this case, we start again, and the expected number of throws until we get a 3 is still \( E(X) \) (since the die has memoryless property).
   So, the expected number of throws in this case is \( 1 + E(X) \).

Now, we can express the expected value of \( X \) using these scenarios: \[ E(X) = \frac{1}{3} \times 1 + \frac{2}{3} \times (1 + E(X)) \] Solving this equation for \( E(X) \): \[ E(X) = \frac{1}{3} + \frac{2}{3}(1 + E(X)) \] \[ E(X) = \frac{1}{3} + \frac{2}{3} + \frac{2}{3}E(X) \] \[ E(X) = \frac{1}{3} + \frac{2}{3} + \frac{2}{3}E(X) \] \[ E(X) = \frac{1}{3} + \frac{2}{3} + \frac{2}{3}E(X) \] \[ \frac{1}{3}E(X) = \frac{1}{3} + \frac{2}{3} \] \[ E(X) = 3 \] So, the expected number of throws until we get a 3 is 3.

Note : if p is the probability of A then expected number of throws to get event A is \( \frac{1}{p} \)