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A bag contains five white balls and five black balls. Your goal is to draw two black balls.

a) You draw two balls at random. What is the probability that they are both black?

 

b) You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

 Feb 7, 2020
edited by AnimalMaster  Feb 7, 2020
 #1
avatar+320 
+1

Okay, I figured out A):

 

FIRST BALL: So the probability of first drawing a black ball or a white ball is both 5/10.

 

SECOND BALL: The probability of getting a second ball is 4 out of 9 (because you already took a ball), so for both it's 4/9.

 

To find the probability, of both, you can multiply 5/10 * 4/9, = 2/9. Which is the answer.

 

I don't know how to do b) though. 

 Feb 7, 2020
 #2
avatar+123 
+4

I agree with your rationale for part a).

 

As for part b), you can apply the same logic of breaking each draw down. For the first two draws, you would have the same probability of \(\frac{2}{9}\) for drawing two black balls. The probability of having one or more white balls at this point is a complement of your desired event of having two black balls, so the probability of having one or more white balls is \(1 - \frac{2}{9} = \frac{7}{9}\). In other words, you will be redrawing \(\frac{7}{9}\) of the time. The probability of drawing two white balls would be the same as drawing two black balls because there are 5 of each, thus P(both white balls) = \(\frac{2}{9}\). Thus, the only other possibility of drawing exactly one white ball and one black ball would be \(1 - \frac{2}{9} - \frac{2}{9} = \frac{5}{9}\)

 

Say you drew one white ball in your first drawing. This happens \(\frac{5}{9}\) of the time. If you put the white ball back, your probability of drawing a second black ball from the 9 with 4 black balls would be \(\frac{4}{9}\), so this probability is \(\frac{5}{9}*\frac{4}{9} = \frac{20}{81}\).

 

Say you drew two white balls in your first drawing. This happens \(\frac{2}{9}\) of the time. If you put both balls back, your probability of drawing two black balls would be the same as the initial drawing, since you have 5 black and 5 white in the bag now. Thus, this probability is \(\frac{2}{9}*\frac{2}{9}=\frac{4}{81}\).

 

Using the sum rule, as these are all disjoint events, and including the probability we chose two black balls on the first draw, we arrive at the probability being \(\frac{2}{9} + \frac{20}{81} + \frac{4}{81} = \frac{14}{27}\). This could be wrong I gotta get to class lmao

 Feb 7, 2020
edited by Anthrax  Feb 9, 2020
 #3
avatar+320 
+2

I think it's a typo, it's 14/27, not 4/27. nice job tho!

AnimalMaster  Feb 7, 2020

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