epicVoid said:
Firm has 10 employees, 2 womens and 8 mens. The Management Board will randomly select 3 employees and reward them with a trip to Venice. What is probability for:
a.) the winners will be only womans
b.) the winners will be only mans
c.) the winners will be exactly 1 women (so other 2 are mans right?)
d.) the winners will be exactly 2 womens (other 1 is man right?)
e). among the winners will be at least 1 woman
solved :
a.) P(A) = 0 (only 2 womans employed)
Obviously. You could also say that there are 2 women out of 10 employees so the probability the first person chosen is 2/10= 1/5. Then there are 1 woman out of 9 employees left so the probability the second person chosen is also a woman is 1/9. Then there 0 women out of 8 employees so the probability the third person chosen is 0/8= 0. The probability all three are women is (1/5)(1/9)(0)= 0.
b.) P(A) = (8 3) / (10 3) = 7/15
Are those binomial coefficients? There are at first 8 men out of 10 employees so the probability the first person chosen is a man is 8/10= 4/5. Then there are 7 men out of 9 employees left so the probability the second person is a man is 7/9. Then there are 6 men out of 9 employees left so the probability the third person is a man is 6/8= 3/4. The probability all three persons chosen are men is (4/5)(7/9)(3/4)= 7/(5)(3)= 7/15 as you say.
c.) P(A) = ((8 2) * (2 1)) / (10 3) -> wrong result
There are 2 women out of 10 employees so the probability the first person chosen is a woman is 2/10= 1/5.There are then 8 men out of 9 employees so the probability the second person chosen is a man is 8/9. There are then 7 men out of 8 employees so the probability the third person chosen is a man is 7/8. The probability of "woman, man, man"
in that order is (1/5)(8/9)(7/8)= 7/45. If you do the same thing with "man, woman, man" and "man, man, woman" you will see that while the fractions are different, the three numerators are the same and the three denominators are the same- so the probability of "man, woman, man" and "man, man, woman" are also 7/45. Since there are three such orders, the probability of "one woman and two men"
in any order is 3(7/45)= 7/9.
d.) have no idea (all I tried was wrong)
e.) have no idea (all I tried was wrong)
There are initially 2 women out of 10 people so the probability the first person chosen is a woman is 2/10= 1/5. There is then 1 woman out of 9 people so the probability the second person chosen is a woman is 1/9. There are then 8 men out of the 8 people so the probability the third person chosen is a man is 8/8= 1 (of course). The probability of choosing "Woman, woman, man"
in that order is (1/5)(1/9)(1)= 1/45. Again, it is easy to see that the probabilities of "woman, man, woman" and "man, woman, man" are also 1/45 so the probability of two women and one man is 3(1/45)= 1/9.
I also tried to calculate c.) like this :
(3 2) * (8/10)^2 * (1-8/10)^1 = 0.384 -> wrong
Thanks for help.
Your arithmetic is wrong!
\begin{pmatrix}3 \\ 2\end{pmatrix}= \frac{3!}{2!1!}= 3
(8/10)^2= .8^2= 0.64
(1- 8/10)^2= .2^2= 0.04
So \begin{pmatrix}3 \\ 2 \end{pmatrix}(8/10)^2(1- 8/10)^2= 3(0.64)(0.04)= 0.0768, not 0.384.
