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A bag contains 3 black balls and 2 white balls. Three balls are selected at random, without replacement.

Edit: Thanks to a commenter below I actually messed up the calculations for the probabilities hence why I wasn't getting the correct answer for any of the questions. I went ahead and fixed the mistakes in each of the questions so that it can help other students.

A) Draw a tree diagram, list the outcomes and their probabilities. I got this answer correct. The outcomes and their probabilities are:

Outcome        Probabilities
BBB                     6/60
BBW                   12/60
BWB                   12/60
BWW                 6/60
WBB                  12/60
WBW                 6/60
WWB                 6/60

B) Find the probability that 2 balls are white.
My Answer: $\frac{6}{60}$ + $\frac{6}{60}$ + $\frac{6}{60}$ = $\frac{18}{60}$ = $\frac{3}{10}$ = 0.30
The balls that I listed are BWW, WBW, and WWB

Edit: I got the correct answer for this one. I made a mistake on the calculations and went ahead and fixed the mistakes for this question.

However the answer key says it's $\frac{3}{10}$ = 0.30. Not sure how they got this answer.

C) Find the probability that 2 balls are white, given that at least one is white.
My answer: $\frac{6}{60}$ + $\frac{6}{60}$ + $\frac{6}{60}$ ÷ $\frac{12}{60}$ + $\frac{12}{60}$ + $\frac{6}{60}$ + $\frac{12}{60}$ + $\frac{6}{60}$ + $\frac{6}{60}$ = $\frac{18}{54}$ = $\frac{1}{3}$ = 0.33

Edit: I got the correct answer for this one. I made a mistake on the calculations and went ahead and fixed the mistakes for this question.

However the answer key says 0.33. It seems that the outcomes for the denominator is correct, but the outcomes for the numerator is incorrect. Not sure where I went wrong.

D) Find the probability that all 3 balls are black, given that at least 2 are black.
My answer: $\frac{6}{60}$ ÷ $\frac{6}{60}$ + $\frac{12}{60}$ + $\frac{12}{60}$ + $\frac{12}{60}$ = $\frac{6}{42}$ = $\frac{1}{7}$ =0.143

Thought behind this is we have BBB ÷ BBB + BBW + BWB + WBB.

Edit: I got the correct answer for this one. I made a mistake on the calculations and went ahead and fixed the mistakes for this question.

However, the answer key says $\frac{1}{7}$= 0.143. Again, I'm not sure where I went wrong. Would appreciate any help I can get for these questions.

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  • $\begingroup$ $P(BWW) = \frac35\frac24\frac13=\frac6{60}$, so $P(2W)=P(BWW)+P(WBW)+P(WWB) =\frac{18}{60}=\frac 3{10}$. $\endgroup$ – user170231 Dec 9 '19 at 22:36
  • $\begingroup$ Thank you so much for responding. I was able to get the correct answers because dumb me made a very silly mistake with the calculations. Thanks again for the help. $\endgroup$ – Cece Dec 9 '19 at 23:10
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BWB 6/60

BWW 12/60

The above are wrong. After drawing one black and one white ball the bag contains more black balls than white. So P(BWB) > P(BWW).

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  • $\begingroup$ Thank you so much for pointing out the mistakes. I actually wrote the correct numbers down but calculated them incorrectly. $\endgroup$ – Cece Dec 9 '19 at 22:37

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