A) 5 to be chosen among a Total : 10 Men + 8 Women
¹⁸C₅ = (18!)/(5!)(13!) = 8,568 groups of five
b) A must to have men and women. If so we have to deduct all groups of 5 that are all men and all group of 5 that are all women
Groups of 5 with only men: ¹⁰C₅ = 252
Groups of 5 with only women: ⁸C₅ = 56
So number of committees of 5 men and women mixed =
8568 - 252 - 56 = 8,260 committees
c) 3 Women and 2 Men:
⁸C₃ x ¹⁰C₂ = 2,520 groups of 3 W and 2 M
d) More women than men, it means:
3 W + 2 M OR (we have found it in c) = 2,520)
4 W + 1 M OR ⁸C₄ x ¹⁰C₁ →→→→ = 700
5 W + 0 M OR ⁸C₅ x ¹⁰C₀ →→→→ = 56
Total where W>M = 3,276 groups of 5 where women are at least 3
Answer:
$260
Step-by-step explanation:
0.06 x 200 = 12
12 x 5 = 60
200 + 60 = 260
Answer: 11+5=16
Step-by-step explanation: 14, 15, 16, 17 and 18
When you have two mutually exclusive events, to find the probability of one or another, you add the probabilities.
65+7=72
Final answer: B
We have that
, which means
reduces to
when
, or reduces to
when
.
So we can expand the summation as
Notice that the 10 contributes a total of 30 from the first sum, and -30 from the second sum, so those terms cancel, leaving us with
