Probability Without Replacement

A bag contains 5 blue balls and 4 red balls. A ball is picked and not replaced. What is the probability of picking at least one red ball?

 

9 balls. 5 are blue and 4 red

P(at least 1 red) = 1 – P(all 5 balls are blue)
P(all 5 balls are blue) =5/9

P(at least 1 red) =1-5/9=4/9=0.4444 => or 44.44%

 

I like probability problems. No more probability for now. Back to precalculus.

 

Are you sure you have copied the problem correctly? It doesn't make sense to say "at least one ball" when you only pick once!
If you pick once then the probability the ball is red is 4/(5+ 4)= 4/9.

"At least one red ball" is the same as "not all blue balls".

If you pick twice, the probability the first ball is blue is 5/9. Then there are 4 blue and 4 red balls so the probability the second ball is also blue is 4/8= 1/2. The probability both balls are blue is (5/9)(1/2)= 5/18 so the probability "at least one ball is red" is 1- 5/18= 13/18.

If you pick three times, the probability both the first and second balls are blue is 5/18 as before. There are then 3 blue balls and 4 red balls so the probability the third ball is also blue is 3/7. The probability the three balls are all blue is (5/18)(3/7)= 5/42 so the probability at least one ball is red is 1- 5/42= 37/42.

Etc.

 

I copied and pasted the problem. I don't know if there is a typo there but this thread is as given. I can't recall where I found this problem. It is not in any of my textbooks. I like your reply. Very informative.