- Messages
- 4
- Reaction score
- 0
- Points
- 1
Thankyou so much!A)
Prob = combinations with tracey and albert/all combinations
Two possible groups according to the rules:
5M1W and 4M2W
all combinations = (10C5)(9C1) + (10C4)(9C2) = 9828
5M1W
One women and one man are chosen, that leaves 4 men to be chosen from 9 men.
That gives us 9C4 combinations
4M2W
One women and one man are chosen, that leaves 3 men to be chosen from 9 men and 1 women to be chosen from 8 women.
That gives us (9C3)(8C1) combinations
Combinations with albert and tracey= 798
Prob = 798/9828
B)
N(either but not both) = N(tracey no albert) + N(albert no tracey)
N(tracey no albert) = N(5M1W with tracey no albert) + N(4M2W with tracey no albert)
N(albert no tracey) = N(5M1W with albert no tracey ) + N(4M2W with albert no tracey )
Using the same logic as in A:
N(tracey no albert) = (9C5) + (9C4)(8C1)
N(albert no tracey) = (9C4)(8C1) + (9C3)(8C2)
N(either but not both) = 4494