5. What computation would you do to figure out for a group of 8 people what the probability is that someone has a birthday of Jan 11? (just write out the mulitplication, don't crunch the numbers)(assume 366 equally likely days per year)

1 - (365/366)*(365/366)*(365/366)*(365/366)*(365/366)*(365/366)*(365/366)*(365/366)

or 1 - (365/366)^8

notice that I have 8 fractions multiplied, because I want each one to not be Jan 11.

6. In a group of 8 people, what is the probability that someone has a birthday in January?

1 - (11/12)*(11/12)*(11/12)*(11/12)*(11/12)*(11/12)*(11/12)*(11/12)

1 - (11/12)^8 = .50

P(a Jan birthday) = 50%

7. If everyone in a goup of 4 people rolls a 6 sided die, what is the probability that at least one person gets a 6?

1 - (5/6)*(5/6)*(5/6)*(5/6)

1 - (5/6)^4 = .52

P(at least one 6) = 52%

8. If I roll a 6-sided die 6 times, what is the probability thta I get at least one 6?

1 - (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)

1 - (5/6)^6 = .67

P(at least one 6) = 67%

9. If everyone in a group of 5 people rolls a 20 sided die, what is the probability that someone will get a 1?

1 - (19/20)*(19/20)*(19/20)*(19/20)*(19/20)

1 - (19/20)^5 = .23

P(at least one 1) = 23%

10. If I roll an 8-sided die 3 times, what is the probability that I get a different number every time?

(7/8)(6/8) or possibly (8/8)(7/8)(6/8)

=.66

P(three different numbers) = 66%

11. If I roll a 12-sided die 4 times, what is the probability that I get a 3 at least once

1 - P(never get a 3)=

1 - (11/12)*(11/12)*(11/12)*(11/12)

1 - (11/12)^4 = .29

P(at least one 3) = 29%

12. If I roll a 6- sided die 5 times, what is the probability that I don't get a 2 on any of the rolls?

(5/6)*(5/6)*(5/6)*(5/6)*(5/6)

(5/6)^5 = .40

P(no 2's) = 40%

13. If I roll a 10 sided die 4 times, what is the probability that I get the same number twice?

1-P(all different)

1 - (9/10)(8/10)(7/10) = .50

P(same number twice) = 50%