math probability?

Question

what is the probability of selecting 2 red socks one after the other out of 20 ?

Answer 1

There is not enough information given.
You need to know how many of the 20 socks are red.

Answer 2

No way to tell. By definition, the probability of one successful event (one draw of a red sock) is P(s = r) = n(r)/N(all); where n(r) is the number of red socks in a pile of N socks of all colors including red and whatever fashion statement you have. You have not given us n(r); that is, how many red socks are intially in that pile of N = 20 socks?

But for grins, let's assume you have n(r) = 2 red socks (a matched pair) in the pile. So, on the first draw P(s = r) = n(r)/N = 2/20 = 1/10. Now on the second draw P(s = r) = n(r)-1/N -1 So the probability of consecutive draws of red is P(s = r AND s = r) = n(r)/N * n(r)-1/N - 1 = 2/10 * 1/19 = 2/190. This presupposes you got a red one on the first draw. If you had not, clearly P(r AND r) = 0. Why? Because without that first red sock you cannot possibly get two in a row. And that's why we use n(r) - 1 for the second draw, one red sock is already gone from the pile, leaving only the one left.

Similarly, we use N - 1 for the second draw because the pile has been reduced by one by the first draw. Anyway, once you figure out how many red socks you have initially in the pile of socks, you can work the problem as indicated.

Answer 3

If you have all red socks, the probability is 1. On the other hand if there are no red socks, the probability is 0. Did you ask the complete question?

Answer 4

Not enough information given in this problem.
I need to know how many red socks there are to choose from the 20

Answer 5

ok.. here's how u do it: first time, it's 2/20 chance, which is 10%. then it's 1/19.. cuz there's 1 left out of 19 possible

So. . 2/20 multiply by 1/19... which is... 2/380 chance.. which is 1/190 chance
that's like... not likely..