Download bability with Combination

ANSWERS TO PROBABILITY REVIEW PROBLEMS
NOTE: I will just write 10P4 and 10C4 for
permutations and combinations; it saves a lot
of typing time.
6. 10C4=210
7. 4P4=4!=24
8. 8C5=56
9. 8P3=336
10. a) 6!/3! = 120
b) 1) 1/6
11. a) 11/36
b) 5/36
12. a) 7 x 6 x 5 = 7P3 = 210
2) 3/6 = ½
3) 3/6 = ½
c) 2/36 = 1/18
b) numbers greater than 400 = 5 x 6 x 5 =150
probability = 150/210= 5/7
13. a) 9C4 = 126
3) 6C4/9C4 = 5/42
b) 1) 3C2 x 6C2/ 9C4 = 5/14
2) 3C1 x 6C3/ 9C4 = 10/21
c) 1 (there aren’t enough girls to make a 4 person committee)
14, a) (4/52) x (4/51) = 4/663
c) (4C1 x 4C1) /52C2 = 8/663
b) 4C2/52C2 = 1/221
d) 13C2/52C2 = 1/17
15. a) 8/52 = 2/13
b) 16/52 = 4/13 (don’t count the 8 of clubs twice)
c) 28/52 = 7/13 (don’t count the 8 of diamonds or 8 of hearts twice)
16. a) P( 3 red)= 3C3/5C3 = 1/10
b) 0 (only two of each color marble)
c) P (3 red) = 3C3/7C3 = 1/35 P(3 blue) = 4C3/7C3 = 4/35
P (3 red or 3 blue) =5/35 = 1/7
d) P(3 white) = 4C3/9C3 = 1/21 P(3 blue) = 5C3/9C3 = 5/42
P( 3 white or 3 blue) = 1/21 + 5/42 = 7/42 = 1/6
e) 1 (all marbles are the same color) f) P(3 blue) = 3C3/ 6C3 = 1/20
17. a) 7C5 = 21 b) 1) (4C2 x 3C3)/ 7C5 = 2/7
2) (4C3 x 3C2)/ 7C5 = 4/7
3) P(at least 2 women) = P ( 3 women 2 men) + P (2 women 3 men) = 2/7 +4/7 =6/7
4) P (all women) = 0 (not enough women to be all women)
c) 5P2= 20
d) This can happen two ways: Hilda chair, Joan sec. OR Hilda sec., Joan chair
Probability = 2 / 20 = 1/10