Two fair dice are tossed, one red one green. Let A be the event that the sum of the numbers on top is 7.?

Question

Let B be the event that the number on top of the red die is not 1. Answer the following:

1) what is the probability that the sum of the numbers on top is 7, i.e. what is P(A)?

2) What is the probability that the sum of the numbers on top is 7, given that the number on top of the red die is not 1, i.e. what is P(A|B)?

Answer 1

These other guys may be wrong on question 2.

The red die is not 1, so 1-6 is eliminated.  but so is 1-1

1-2 1-3 1-4  and 1-5

So the answer is 5/30 = 1/6 = 16.67%

Answer 2

1-6

2-5

3-4

4-3

5-2

6-1

1) P(A) = 6/36 = 0.167

2) P(A/B) = 5/36 = 0.139

Answer 3

Are you talking about this big oversized fair dice? The odds of winning is practically zero, unless you're working for them. Those games are set up to take your money, not to give suff away. I've seen games that seem easy as all h**l, in theory, and a person be one try away from winning and end up spending hundreds of dollars all for a stupid stuffed animal they could have bought from Walmart for $20 or less. Your best bet is to watch other people play for a while. Once you figure out those statistical odds, if you notice people win MUCH less than statistics say, they're probably trying to s***w you..otherwise, give it a go.

Answer 4

Do you mean Fair, as in not weighted.

Answer 5

<<<1) what is the probability that the sum of the numbers on top is 7, i.e. what is P(A)?>>>

6 out of 36 = 1 out of 6 = 0.1667 = 16.67%

<<<2) What is the probability that the sum of the numbers on top is 7, given that the number on top of the red die is not 1>>>

This combination of (red 1 + green 6) has been elimintated leaving only 5 ways to make a 7, so the probability is

5 out of 36 = 0.1389 = 13.89%