help needed. thank you!

Question

1. (Question not related to a deck of 52 cards.) You have 5 blue cards, 4 red, and 6 green in a box. If you draw two cards from the box with replacements, find P(Blue and Green).
2. Two six sided die are rolled. What is the probability that the sum will be less than 10?
3. A group consists of 3 people. If 11 people are able to be in a group, how many different groups can be made?

Answer 1

4 Red, 5 Blue, 6 Green, out of 52 cards total

P(B) = 5/52, P(G) = 6/52 = 3/26

P(B&G) = P(B)xP(G) + P(G)xP(B) = 2P(B)xP(G) = 2(5/52)(3/26) = 30/(52)(26) = 15/52(13) = 15/676 = about 2.2%

2 one way 1&1

3 two ways 2&1, 1&2

4 three ways 3&1, 1&3, 2&2

5 four ways

6 five ways

7 six ways

8 five ways

9 four ways

10 three ways 6&4, 4&6, 5&5

11 two ways 5&6, 6&5

12 one way 6&6

total 36 ways

less than 10 happens in 30 ways

P(<10) = P(2,3,4,5,6,7,8 or 9) = 1-Pr(10,11,or 12) = 1 -6/36 = 30/36 = 5/6 = about 83.3%

11C3 = 11!/3!8! = 11(10)(9)/3(2)(1) =11(5)(3) = 161 combinations

if order matters within each group then

11P3 = 11/8! = 11x10x9 = 990 permutations

for combinations ABC, ACB, BAC, BCA, CAB, CBA are all counted as just one group

for permutations they count as 6 different groups

Answer 2

#1 There are 2 possible outcomes to meet the criteria: BG and GB. Write the probability of each, and since they are independent events, add.

#2 Write a 6 x 6 table with the sum of each row and column at each intersection. Then count the intersections which meet the criteria,

#3 This is a straight application of binomial coefficients.