Answer:
Let's define the following events:
- A: the event that a prime number is rolled.
- B: the event that an even number is rolled.
- C: the event that a one is rolled.
The probability of rolling a prime is 3/6 (since three sides of the die have prime numbers). Similarly, the probability of rolling an even number is 3/6. The probability of rolling a one is 1/6.
We want to find the expected value of the number of ones rolled, which we can do by finding the expected number of rolls until either two primes or two events are rolled. Let E be the expected number of rolls.
There are two cases to consider: either the first two rolls are both ones, or they are not.
If the first two rolls are both ones, then we have already rolled one of the two types of numbers we need, so we just need to keep rolling until we get the second type. The probability of this happening is (3/6)^2 = 1/4, and the expected number of rolls, in this case, is E+2 (since we have already rolled two ones).
If the first two rolls are not both, then we need to keep rolling until we get two primes or two events. Let P be the expected number of rolls until this happens.
If the first roll is one, then the probability of the second roll being a prime or even is 3/5 (since there are three primes or evens left out of five possible outcomes). In this case, we would need to roll one more time to get the second number of that type, so P = 1 + (3/5)E.
If the first roll is prime or even, then the probability of the second roll being the same type of number is 2/5. In this case, we would need to roll one more time to get the second number of that type, so P = 1 + (2/5)P. Solving this equation gives P = 2.5.
So we have E = 1/4(E+2) + 3/4(2.5), which simplifies to E = 7/3.
Therefore, the expected number of ones rolled is E-2 = 1/3.
Explanation: