Bob rolls a fair six-sided die each morning. If Bob rolls a composite number, he eats sweetened cereal. If he rolls a prime number, he eats unsweetened cereal. If he rolls a 1, then he rolls again. In a non-leap year, what is the expected number of times Bob will roll his die?
Let the expected value of the number of times Bob rolls his die on a single day be $E$. When Bob rolls his die, there is a $\frac{5}{6}$ chance that he will stop rolling after one roll, and a $\frac{1}{6}$ chance that he will have to start over. In the second case, since his first roll had no effect on the outcome, Bob will on average roll his die $E$ more times, for a total of $1+E$ times that day. Therefore, we know that $E=\frac{5}{6}(1)+\frac{1}{6}(1+E)$, or that $E=\frac{6}{5}$. Over 365 days, Bob will roll his die an average of $\frac{6}{5}\cdot365=\boxed{438}$ times total.
438
Level 5
Counting & Probability
Problem: Bob rolls a fair six-sided die each morning. If Bob rolls a composite number, he eats sweetened cereal. If he rolls a prime number, he eats unsweetened cereal. If he rolls a 1, then he rolls again. In a non-leap year, what is the expected number of times Bob will roll his die?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the expected number of times Bob will roll his die in a non-leap year, we first need to analyze the outcomes of a single roll of the die. The die has six faces, numbered 1 through 6. The composite numbers among these are 4 and 6, the prime numbers are 2, 3, and 5, and the number 1 is neither prime nor composite.

When Bob rolls a 1, he must roll again. This introduces a recursive element to the problem. We denote the expected number of rolls as E. The probabilities of rolling each number are equal, each being 1/6.

The outcomes can be summarized as follows:
- Rolling a composite number (4 or 6) occurs with a probability of 2/6, and Bob stops rolling.
- Rolling a prime number (2, 3, or 5) occurs with a probability of 3/6, and Bob also stops rolling.
- Rolling a 1 occurs with a probability of 1/6, and Bob rolls again.

We can express the expected number of rolls E as:
E = (2/6)(1) + (3/6)(1) + (1/6)(E + 1)

Breaking this down:
- The first term (2/6)(1) accounts for the cases where Bob rolls a composite number and stops after one roll.
- The second term (3/6)(1) accounts for the cases where Bob rolls a prime number and also stops after one roll.
- The third term (1/6)(E + 1) accounts for the case where Bob rolls a 1, which requires him to roll again, thus adding 1 to the expected rolls.

Now, we simplify the equation:
E = (2/6) + (3/6) + (1/6)(E + 1)
E = 5/6 + (1/6)E + (1/6)

Combining like terms gives:
E = 1 + (1/6)E

To isolate E, we rearrange the equation:
E - (1/6)E = 1
(5/6)E = 1
E = 1 / (5/6)
E = 6/5

Now, we have the expected number of rolls for a single day as 6/5. In a non-leap year, there are 365 days. Therefore, the total expected number of rolls in a year is:
Total Expected Rolls = 365 * (6/5) = 438.

Thus, the final answer is 438.<|END_OF_LATENT|> The final answer is $438$.