You roll two fair dice. Let E be the event that an even total shows on the dice. Let F be the event that a six shows on at least one of the dice.

Find P(F) and P(F|E).

Choices:

P(F)=1/3; P(F|E)=1/13
P(F)=1/3; P(F|E)=5/18
P(F)=11/36; P(F|E)=7/18
P(F)=11/36; P(F|E)=5/18?

Would the answer be,

P(F)=11/36; P(F|E)=5/18?

Agree.

Oh, I see you're talking about dice probabilities. Let's roll with it and find the correct answer!

To find P(F), the probability that a six shows on at least one of the dice, we can use the complementary probability. In other words, we'll find the probability that no sixes show up on both dice and subtract it from 1. Since each die has 6 equally likely outcomes, the probability of not rolling a six on one die is 5/6. Therefore, the probability of not rolling a six on both dice is (5/6) * (5/6) = 25/36. Subtracting this from 1 gives us P(F) = 1 - 25/36 = 11/36.

Now, to find P(F|E), the probability of rolling a six on at least one die given an even total, we need to consider the outcomes that satisfy both events. There are a total of 18 outcomes that result in an even total (2, 4, 6, 8, 10, 12) out of a possible 36 outcomes for rolling two dice. Out of those 18 outcomes, 3 of them have a six on at least one die (66, 46, 64). Therefore, P(F|E) = 3/18 = 1/6.

So, the correct answer would be P(F) = 11/36 and P(F|E) = 1/6. I hope this clarifies things for you!

To find the probabilities, we first need to determine the sample space and the desired outcomes.

The sample space for rolling two fair dice consists of 36 equally likely outcomes, as there are 6 possible outcomes for each die.

To find P(F), the probability that a six shows on at least one of the dice, we need to find the number of outcomes where at least one of the dice shows a six. There are 11 such outcomes: {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. Thus, P(F) = 11/36.

To find P(F|E), the conditional probability that a six shows on at least one of the dice, given that an even total shows on the dice, we need to find the number of outcomes where both conditions are met (i.e., even total and at least one six) and divide it by the number of outcomes where an even total shows on the dice.

For an even total to show on the dice, the possible outcomes are {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4)}. Out of these, the outcomes where at least one six shows on the dice are {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

Therefore, P(F|E) = 11/16.

Hence, the correct choice is P(F) = 11/36; P(F|E) = 11/16.

To find the probabilities, we first need to determine the total number of outcomes and the number of favorable outcomes for each event.

There are 36 possible outcomes when rolling two fair dice since each die has 6 possible outcomes (1, 2, 3, 4, 5, 6) and there are two dice (6 × 6 = 36).

Now let's find the number of favorable outcomes for event F, the event that a six shows on at least one of the dice.

There are 11 outcomes where a six shows on at least one of the dice: (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), and (6, 6). Since there are 11 such outcomes, the probability of event F occurring is P(F) = 11/36.

Next, let's find the number of favorable outcomes for event E, the event that an even total shows on the dice.

There are 18 outcomes where an even total shows on the dice: (2, 1), (4, 1), (6, 1), (2, 2), (4, 2), (6, 2), (2, 3), (4, 3), (2, 4), (4, 4), (6, 4), (2, 5), (4, 5), (6, 5), (2, 6), (4, 6), (6, 6). Since there are 18 such outcomes, the probability of event E occurring is P(E) = 18/36 = 1/2.

Finally, to find the conditional probability P(F|E), we divide the number of outcomes where both events E and F occur by the number of outcomes where event E occurs.

The outcomes where both events E and F occur are (2, 6), (4, 6), and (6, 6), so there are 3 such outcomes.

Therefore, P(F|E) = 3/18 = 1/6.

So the correct answer is indeed P(F) = 11/36; P(F|E) = 1/6.

The closest choice is P(F) = 11/36; P(F|E) = 5/18, but it is not the correct answer.