Recent content by theone

yes, it does. Thanks So then the reason that you can't say P(N=1 | N+S=2) = (sum of the "1" row) * (sum of the "1" column) is because those two probabilities being multiplied are from a different "sample space", the one where N+S is not equal to 2 ? Thanks, I think the problem is that it's...

would that be 1/3? In the solution, I am seeing that the way to get P(N=1 | N+S=2) is to do P(N=1 | S=1)/P(N+S=2) = 0.18/0.40 = 0.45 but I do not understand the reason for dividing the probabilities and the first number should be 0.04, I will edit it

Homework Statement Let N denote the number of accidents occurring during one month on the northbound side of a highway and let S denote the number occurring on the southbound side. Suppose that N and S are jointly distributed as indicated in the table. N/S 0 1...

you're right, that's the mistake. I am now getting r=8 as I should I am also confused about the meaning of independent vs dependent. Does all it mean is whether you are replacing after each selection or not? So if its was independent, would it be: 4 (r/27)(27-r/27)(27-r/27)(27-r/27) ?

With P(one of the damaged items is insured) = 4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24) P(none of the damaged items are insured) = (27-r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24) I end up with r=9 But according to the solution, I should be getting r=8

I found a solution online and it takes the r/27 approach so I think have to go with that P(one of the damaged items is insured) = 4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24) P(none of the damaged items are insured) = (27-r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24) For the first probability...

Homework Statement From 27 pieces of luggage, an airline handler damages a random sample of 4. The probability that exactly one of the damaged items is insured is twice the probability that none of the damaged pieces are insured. Calculate the probability that exactly two of the four damaged...

to get the probability of x falling within a range, aren't you essentially adding the probabilities of x taking on the particular values within the range... but if the probability of x taking on a particular value is 0, then why is this sum not always zero?

the probability that a continuous random variable X takes on one of its possible values x?

Homework Statement Can someone explain why f(x) = 1/(b-a) for a<x<b ? Homework EquationsThe Attempt at a Solution shouldn't it be 0? since its a continuous random variable and so that interval from a to b has an infinite number of possible values?

Do you mean this: ##E(X) = \sum_{i} X_i P(X_i)##

the expected value of the random variable X; the probability weighted average of the possible outcomes of X

Homework Statement Does anyone know of a simple proof for this: https://s30.postimg.org/tw9cjym9t/expect.png E(X) = E(X|S)P(S) + E(X|S_c)P(S_c) X is a random variable, S is an a scenario that affects the likelihood of X. So P(S) is the probability of the scenario occurring and and P(S_c) is...

Homework Statement attached picture Homework EquationsThe Attempt at a Solution I read about the virtual condition http://www.symphonytech.com/articles/GDT_Design.htm but I don't really understand the approach used to solve the problem and was wondering if someone could explain the solution...

is it just the vertical contact forces between the tire/ground and the trailer tongue/ground as well as the horizontal friction force from the tires? The full question involves the design of a dolly for the trailer. So when the dolly is connected to the trailer, should that just be taken as...