Thread: some basic-ish probablity questions

Hi,

Just doing some practice questions on probability and got stuck on a few of them:

Q1: Roll one die. Stop a the 4th roll or when a 6 appears (whichever occurs first). Let X be the number of rolls. Find the pmf, expectation and variance of X.

I know how to calculate these things. It's just formulating a table of values that I'm struggling with here so any advice would be welcome.

Q2: To check the quality if drainpipes a manufacturer slects 10 pipes at random from each large production batch and subjects them to an impact test. Flawed pipes break on the test but good ones don't. If more than 2 pipes on the test break, the whole batch is scrapped.
What is the probablity that the batch will be scrapped when 10% of the pipes in it are flawed?

Here i know that P(batch is scrapped) = 0.2
Is this therefore some sort of conditional probability question? Any hints to get me started again would be welcome

Thanks

It's just formulating a table of values that I'm struggling with here so any advice would be welcome.

Your column of X is the number of rolls it takes, your column of P(X) is the probability of getting a 6 on that roll:

X | P(X)
1 1/6
2 (5/6)*(1/6)

. . .

Do you see why I multiply by 5/6 on the second roll. The probability of rolling twice, is the probability that you DIDNT get a 6 on the first roll, and you got one on the second.

X=4 is a tricky one. Try to use some reasoning when you get to that point; because if you reason incorrectly, and add up your probabilities - you will NOT get a 1 (or 100%).

Here i know that P(batch is scrapped) = 0.2

Close, but flawed reasoning. This is a binomial formula problem, with P being the pipe breaks, and Q beign the pipe doesnt break. The probability that that pipe breaks is 0.1 (do you see why?). The probability that it doesn't is simply the compliment of that. The probability that one pipe breaks on one test is independent of the next pipe breaking. Therefore we are conducting 10 independent tests, with a probability of success of 0.1 on any one test.

Thanks for that. My values for P(x) in the first question all add up to one.
For P(4) you have (5/6)^3 * (1/6) + (5/6)^4 as the game ends irrespectively

That is correct.