A raffle held by Mundane Products has collected 700 entries. They are giving away 40 TVs and 30 stereos. John bought 5 entries for the raffle.

What is the probability that John wins exactly one TV as well as exactly one stereo?

John just heard that the TVs are already decided and he has won exactly one of the 40 TVs. What is the probability that John will win at least one stereo in the remaining drawing?

To find the probability of John winning exactly one TV and exactly one stereo, we need to consider the total number of ways he can win both prizes and divide it by the total number of possible outcomes.

First, let's calculate the probability of John winning exactly one TV and exactly one stereo.

The number of ways John can win one TV is 1 (since he already won one) out of the total number of TVs being given away, which is 40. So, the probability of John winning one TV is 1/40.

Similarly, the number of ways John can win one stereo is 1 out of the total number of stereos being given away, which is 30. So, the probability of John winning one stereo is 1/30.

To find the probability of both events occurring, we multiply the probabilities together:

Probability of winning exactly one TV and exactly one stereo = (1/40) * (1/30) = 1/1200.

Therefore, the probability that John wins exactly one TV as well as exactly one stereo is 1/1200.

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To find the probability of John winning at least one stereo in the remaining drawing, we need to consider the total number of ways he can win at least one stereo and divide it by the total number of possible outcomes.

Since John has already won exactly one TV, there are now 39 TVs remaining.

To find the probability of John winning at least one stereo, we need to calculate the probability of not winning any stereos and subtract it from 1. In other words, we want to find the probability that John wins no stereos and then subtract that from 1 to get the probability of winning at least one stereo.

The number of ways John can win no stereos is the total number of remaining stereos (30) minus the number of ways he can win one stereo (1), which is 29.

So, the probability of John not winning any stereos is 29/30.

Therefore, the probability of winning at least one stereo is 1 - (29/30) = 1/30.

Thus, the probability that John will win at least one stereo in the remaining drawing is 1/30.