Math Problem

LameR · Post by **LameR** » 05 Jun 2005 23:08

Got a final in 2 days, and my friend is taking it tomorrow. I can find out how to do it from classmates tomorrow, but she can't because...she doesn't have another day. So if you can help out, it'd be much appreciated. I think it's pretty simple, but I forgot how to do it.

There are 5 red and 4 black balls ina box. IF 3 balls are picked without replacement, what is the probabiliity that at least one of them is red?

wolfpac444 · Post by **wolfpac444** » 05 Jun 2005 23:18

Okay. For these "at least" problems, it is often easier to calculate the probability that you get no red balls. So, here's how it goes.

For the first draw, the probability that you pick a black ball is 4/9.

For the second draw, the probability that you pick another black ball is 3/8 (since you already took a black ball out of the box).

Finally, the probability for choosing a black ball the third time is 2/7.

Therefore, the probability of drawing all black balls is ( 4/9 )*( 3/8 )*( 2/7 )=3/63 = 1/21 . Thus, the probability of drawing at least one red ball is 1 - 1/21 = 20/21.

LameR · Post by **LameR** » 05 Jun 2005 23:23

Thanks a ton Mike!

bigdirtyfoot · Post by **bigdirtyfoot** » 06 Jun 2005 04:00

Wow, this reminds me of how much I really hated AP Statistics.