Bignose

One of the fluid mechanics books on my shelf (by Munson, Young, and Okiishi) says that if you use mercury in a barometer, the vapor pressure of mercury at 68 deg F is 0.00023 lbf/in^2. For engineering purposes, this is pretty close to zero. For most any fluid, the vapor pressure versus temperature can be looked up in an appropriate handbook. Most have been fitted to a polynomial: P_vap(T) = A + B*T + C*T^2 + D/T The handbook will have the values of A,B,C & D, as well as the range over which this polynomial fit is valid.

This is a question that really should receive more attention. Look up the Eddy Curry situation with the basketball team the Chicago Bulls. Curry has a heart defect, and the Bulls wouldn't sign him to an expensive contract. The kicker is that many people have the same defect Curry does, go on to have long fruitful lives. In fact, most people who have the defect probably don't even know they have it, since it is rare that it presents itself. It is a tough question -- do you pay someone several million dollars of a guaranteed contract when there is a significant chance he won't even be able to play? On the other hand, it is much more likely that he will be completely fine and by denying him it is akin to discrimination. This question wasn't ever really answered since the Bulls traded him to the New York Knicks, but as technology explores more of these questions, they are going to come up more and more often.

The definition of independent probabilities is that irregardless of the conditions in that probability, the probability remains the same. In symbols. Events A & B are independent if P(A|B) = P(A) and P(B|A) = P(B). Successive flips of a (fair) coin are independent. That is, it does not matter how many heads there has been in a row, the next flip P(H) = 0.5 and P(T)=0.5. Next time you go to the casino you can no confidently make fun of the people who bet on the roulette table "Oh, there have been 4 blacks in a row, so red MUST be coming up." Sampling of the balls in your example can be independent if you replace the drawn out ball after each pick. With replacement, the probabilities remain 4/7, 2/7, and 1/7 for R,B,G respectively. The conditional probability in my 1st response has to be used since you gave additional information that the ball drawn out is not red. So, you have to look at the subpopulations with the given information that the drawn ball is not red. If you sample the balls without replacement, then once again you have to start using conditional probablilties. For example, we can calculate all the probabilites if the 1st ball is red. If the 1st is R, then the remaining population is 3R, 2B, 1G. So, the probabilites in order are 3/6=1/2, 2/6=1/3 and 1/6. so Probability of drawing a red ball given that the 1st drawn ball is red= P(R|R) = 1/2 P(B|R) = 1/3 P(G|R) = 1/6 and so on for all the cases. Finally, a lot of these things are explained in introductory probability books. There are a lot of good ones out there, with lots more examples than just this. I'd encourage you to go check them out if you want to understand this al lot more fully.

In order to calculate the "not" quantities, you have to set up the conditional probabilities correctly. You got started on the right foot. The Probability of Not Red = P(NR) is 3/7. Now, you have to look at the remainder of the distribution: Probability of Blue given that the drawn ball is Not Red = P(B | NR) = 2/3 which is also equal to Probability of Not Green given Not Red = P(NG | NR) Probability of Green given Not Red = P(G | NR) = P(NB | NR) = 1/3 The vertical bar is read "conditional that", that is P(G | NR) is probability of green given that the ball is not red. Now, if are given that the ball is not Red and Not Blue, you have to look at the remaining subpopulation. P(G | NR & NB) = 1 since there is only a green ball left. Now, the total Probability of green is P(G | NR & NB)*P(NB | NR)*P(NR) = 1*(1/3)*(3/7) = 1/7 just as calculated in the first place.