Balls in a Bag - Difficult Twist (for me)

Badger

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There are 25 balls in a bowl. 7 black, 6 blue, 5 green, 4 red, and 3 yellow. Balls must be drawn until 4 different colors are drawn. What is the probability that one is black?

Please walk me through if you can. Thank you.
 
Have you considered that it might be substantially easier to determine the probability that NONE is black?
 
There are 25 balls in a bowl. 7 black, 6 blue, 5 green, 4 red, and 3 yellow. Balls must be drawn until 4 different colors are drawn. What is the probability that one is black?
This does not make sense. The part in blue is in conflict with the part in red. Please edit the question or post a correction.
 
That was my initial reaction, too, but I decided it was marginally okay the way it stands, albeit confusing.

There are five colors in the bag. If there were only four colors, the probability would be one (1).
"one is black" means "at least one is black".

My views. I welcome others'.
 
Hello, Badger!

This problem is tricky to read.
And its solution is even trickier . . . No, I don't have it.


There are 25 balls in a bowl: 7 black, 6 blue, 5 green, 4 red, and 3 purple.
Balls must be drawn until 4 different colors are drawn.
What is the probability that exactly one is black?

Balls are drawn one at a time, without replacement, until five colors are obtained.

The outcomes could range from 5 draws to 23 draws.


Some examples:
. . . . . . . . . . . . . . . . . X X X X X[c. . .
and its permutations

. . . . . . . . . . . . . . . . . . . . \(\displaystyle \vdots\)

. . . .
X X X X X X X X X X X X X X X X X X X X X X X. .
and the permutations of the first 22 balls


When all the above outcomes have been counted,
. . consider: How many of them have exactly one black ball?
 
Last edited:
I apologize for the lack of specificity. Yes, I should have said "at least one ball is black." I'm guessing there is no elegant equation that would answer this and that the only way is to work out all the possible permutations?
 
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