I also tried to calculate c.) like this, but get wrong result :
(3 2) * (8/10)^2 * (1-8/10)^1 = 0.384 -> wrong
I have a homework which can not be fully resolved with my knowlage:
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)
b.) P(A) = (8 3) / (10 3) = 7/15
c.) P(A) = ((8 2) * (2 1)) / (10 3) -> wrong result
d.) have no idea (all I tried was wrong)
e.) have no idea (all I tried was wrong)
Thanks for help.
I also tried to calculate c.) like this, but get wrong result :
(3 2) * (8/10)^2 * (1-8/10)^1 = 0.384 -> wrong
You answer for c looks fine to me. How do you know it's wrong?
I don't have emotions and sometimes that makes me very sad.
Our teacher gave us results:
a.) 0
b.) 7/15
c.) 7/15 = 0.467
e.) 8/15 = 0.533
d is made up by me, so I don't have a result.
I calculated all again and get this :
a.) 0 // teacher's result = 0
b.) (3 3) * (8/10)^3 * (1-8/10)^0 = 1 * 512/1000 * 1 = 0.512 // teacher's result = 7/15 = 0.467
c.) (3 2) * (8/10)^2 * (1-8/10)^1 = 3 * 64/100 * 2/10 = 0.384 // teacher's result = 7/15 = 0.467
d.) (3 1) * (8/10)^1 * (1-8/10)^2 = 3 * 8/10 * 4/100 = 0.096 // unknown result
e.) c.) + d.) = 0.384 + 0.096 = 0.480 // teacher's result = 8/15 = 0.533
All worng except a.)...
Problem c.) There are three possible ways to end up with exactly 1 woman in three selections: (M and M and W) or (M and W and M) or (W and M and M). The associated probabilities are P(M&M&W) = (8/10)×(7/9)×(2/8) = 14/90, P(M&W&M) = (8/10)×(2/9)×(7/8) = 14/90 and P(W&M&M) = (2/10)×(8/9)×(7/8) = 14/90. Any of these three outcomes will get the desired end result, and so the probability of ending up with exactly 1 woman is P(1W) = (14/90)+ (14/90)+ (14/90) = 42/90 = 7/15.
Problem d.) There are three possible ways to end up with exactly 2 women in three selections: (M and W and W) or (W and M and W) or (W and W and M). The associated probabilities are P(M&W&W) = (8/10)×(2/9)×(1/8) = 1/45, P(W&M&W) = (2/10)×(8/9)×(1/8) = 1/45 and P(W&W&M) = (2/10)×(1/9)×(8/8) = 1/45. Any of these three outcomes will get the desired end result, and so the probability of ending up with 2 women is P(2W) = (1/45)+ (1/45)+ (1/45) = 3/45 = 1/15.
Problem e.) There are two ways to end up with at least one woman: 1W or 2W. The associated probabilities are P(1W) = 7/15 and P(2W) = 1/15. Any of these two outcomes will get the desired end result, and so the probability of ending up with at least 1 woman is P(≥1W) = 7/15+1/15 = 8/15.
The above methods show the value of understanding probability problems in terms of first principles (pigeonholing & counting), as well as the mathematical meaning of “and” and “or” when calculating the probabilities.
epicVoid (05-23-2014)
Thank you. I understand now.
I don't have emotions and sometimes that makes me very sad.
If I calculate ((8 2) * (2 1)) / (10 3), I get :
(28 * 2) / 4320 = 0.00129...
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