probabilitysolution-verification
I'm looking for confirmation that what I for did this question is correct:
"If two people are randomly chosen from
a group of eight women and six men, what is the probability
that (a) both are women (b) one is a man and the other is a
woman?"
a) The probability that both are women: $\frac{8}{14}\times\frac{7}{13}\approx.31$
b) The probability that one is a man and the other is a woman: $\frac{6}{14}\times\frac{8}{13}\approx.26$
(1) Your answer is correct. It appears you might be using software instead of normal tables to get so many decimal places of accuracy. [To use normal tables, you would have to 'standardize' (convert to standard normal distributions), then get something like four digits of accuracy.] In R software, this computation is as follows, without standardizing.
qnorm(.25, 69.3, 2.8)
## 67.41143
1 - pnorm(67.4114, 64, 2.7)
## 0.1032081
In the graph below, 25% of the probability under the blue curve (for men, at right) lies to the left of the dashed line at 67.4114, and 10.32% of the probability under the orange curve lies to the right of the same vertical line. (I recommend that you always try to draw sketches for such problems, especially as problems become more intricate than this one. Even very rough sketches can help catch gross computational or logical errors.)
(2) Let $X$ be the height of a randomly chosen woman and $Y$ be the height of a randomly chosen man. This part requires you to look at the distribution of the difference $D = Y - X$. Then $D$ is normally distributed with $$E(D) = \mu_D = \mu_M - \mu_W = 69.3 - 64 = 5.3$$ and $$V(D) = \sigma_D^2 = \sigma_M^2 + \sigma_W^2.$$ Notice that you subtract the means and add the $variances$. (So far, you have been dealing with standard deviations.) Then you want $P(D > 5.3).$ From what you have shown, I don't think you should have trouble from there on. (Make a sketch. Even without computations, the answer should be obvious.)
I don't know if you care for simulations, but here are results of a million simulated performances of this 2-person experiment. Simulated results are not perfectly accurate, but you can use them as a 'reality check' on your work.
x = rnorm(10^6, 64, 2.7)
y = rnorm(10^6, 69.3, 2.8)
d = y - x
mean(d); sd(d); mean(d > 5.3)
## 5.302365 # approx E(D)
## 3.889633 # approx SD(D)
## 0.50055 # approx P(D > 5.3)
(3) This part is very similar to part (2), but the result is not obvious, and you have a little computation to do. (My simulated answer is nearer to 0.53 than to 0.54.)
If this does not put you on the right track, or if you have unresolved questions, please leave a Comment.
Let's call the "first" position treasurer and "second" position secretary. Then our order of choice matters in this setup.
Your initial computation (9/35) is correct since it computes the probability of putting a man in the first position and a woman in the second.
Your next computation (18/35) computes the ratio of the number of ways to choose a man and woman to the total number of ways to choose two people. But the numerator must be halved to account for the fact that only half of those ways place the man in the first position and woman in the second. This results in the correct answer.
Best Answer
The first one is correct.
However, to see the problem with part $b$ try to answer this question: What is the probability that the first person is men and the second one is a woman ?