Probability of pulling a black ball

B

bry32321

New member
Joined
Aug 23, 2013
Messages
23
A big box contains 3 black balls and 7 white balls. In Step 1, half of the balls are taken out of the big box (without replacement) and put in a small box, which was originally empty.
(a) For each x = 1, 2, 3 Find the probabilities that x black balls and 5-x white balls were transferred

My work:

P(x) = ((3 choose x)(7 choose 5-x))/(10 choose 5)

P(x=1) = 0.417

P(x=2) = 0.417

P(x=3) = 0.083

(b) In Step 2, one ball is then drawn at random from the small box. Find the probability that this ball is black, if x black balls and 5-x white balls were transferred, for each x = 1, 2, 3.

My work:
x=1 means there are 4 white balls and 1 black ball so the probability of the ball being black when pulled is 1/5 = 0.2

x=2 means there are 3 white balls and 2 black balls so the probability of the ball being black when pulled is 2/5 = 0.4

x=3 means there are 2 white balls and 3 black balls so the probability of the ball being black when pulled is 3/5 = 0.6

(c) In part (b) suppose the randomly drawn ball is observed to be black. What is the probability that 2 black balls and 3 white balls were transferred from the big box to the small box?

My work:

I was thinking maybe I should consider the different combinations of 2 black and 3 white so I did this:
BBWWW
BWBWW
BWWBW
BWWWB
WBWWB
WWBWB
WWWBB
WWBBW
WBBWW
WBWBW

But after that I am lost.



Are these anywhere near correct? I would really appreciate any feedback and help that I can get on these problems. Thank you!
 
bry32321 said:
A big box contains 3 black balls and 7 white balls. In Step 1, half of the balls are taken out of the big box (without replacement) and put in a small box, which was originally empty.
(a) For each x = 1, 2, 3 Find the probabilities that x black balls and 5-x white balls were transferred

My work:

P(x) = ((3 choose x)(7 choose 5-x))/(10 choose 5)

P(x=1) = 0.417

P(x=2) = 0.417

P(x=3) = 0.083
They didn't ask for it, but it is interesting to note that
.....P(x=0) = 0.083
.....P(x>1) = 1 - 0.083 = 0.917

(b) In Step 2, one ball is then drawn at random from the small box. Find the probability that this ball is black, if x black balls and 5-x white balls were transferred, for each x = 1, 2, 3.

My work:
x=1 means there are 4 white balls and 1 black ball so the probability of the ball being black when pulled is 1/5 = 0.2

x=2 means there are 3 white balls and 2 black balls so the probability of the ball being black when pulled is 2/5 = 0.4

x=3 means there are 2 white balls and 3 black balls so the probability of the ball being black when pulled is 3/5 = 0.6
Write these as conditional probabilities:
......P(black | x=1) = 0.2
......P(black | x=2) = 0.4
......P(black | x=3) = 0.6

(c) In part (b) suppose the randomly drawn ball is observed to be black. What is the probability that 2 black balls and 3 white balls were transferred from the big box to the small box?

My work:

I was thinking maybe I should consider the different combinations of 2 black and 3 white so I did this:
BBWWW
BWBWW
BWWBW
BWWWB
WBWWB
WWBWB
WWWBB
WWBBW
WBBWW
WBWBW

But after that I am lost.

Are these anywhere near correct? I would really appreciate any feedback and help that I can get on these problems. Thank you!
Click to expand...
For part (c), try combining the conditional probabilities to get the total probability of drawing black:

\(\displaystyle \displaystyle P(\text{black}) = \sum_{x=1}^3 \left[P(\text{black}\ |\ x) \times P(x)\right]\)

Of all those ways to get "black," what fraction is from the x=2 term?
 
For part c, I first found the probability of black being drawn using the values from part a and b so I got,

P(black drawn)=(1/5)(.417)+(2/5)(.417)+(3/5)(.083)=0.3

Then, I found the probability of x=2 given a black ball is drawn,

P(x=2│black drawn)=P( black drawn union x=2)/P(black drawn)= (P(x=2)P(black drawn│x=2))/P(black drawn)= ((.417)(2/5))/(3/10)= 0.556

And thank you for your help!
 
Last edited:
You must log in or register to reply here.
Top