+569 A bag contains five white balls and four black balls. Your goal is to draw two black balls.
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.)
I have tried, but I am stuck. :( help.
First way: [5/9 * 4/8 + 4//9 * 3/8] = 4/9
2nd way: [9C2 ] = 36 number of ways you can 2 balls
Number of ways of drawing 2 balls of the same color =5C2 + 4C2 =16
Probability =16 / 36 = 4/9
+118667 Attn: Iamhappy
I do not like that you have given your guest answerer thumbs down.
If someone answers you, you should thank them for their interest and their time.
If you do not think the answer is correct then you open a polite conversation with your reasons.
To give thumbs down to someone who has tried to help you is rude.
Note: I have not looked at the answer as yet so I have no idea whether or not it has any merit.
But thank you guest answerer for your time and interest.
+118667
A bag contains five white balls and four black balls. Your goal is to draw two black balls.
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.)
Winning draws
BB (4/9)*(3/8) = 12/72
BW B (4/9)*(5/8)*(3/8) = 60/576
WB B (5/9)*(4/8)*(3/8) = 60/576
WW BB (5/9)*(4/8)*(4/9)*(3/8) = 240/5184
ADDED = 12/72+120/576+240/5184 = 91/216
Non winning draws
WW WB (5/9)*(4/8)*(5/9)*(4/8) = 400/5184
WW BW (5/9)*(4/8)*(4/9)*(5/8) = 400/5184
WW WW (5/9)*(4/8)*(5/9)*(4/8) = 400/5184
WB W (5/9)*(4/8)*(5/8) = 100/576
BW W (4/9)*(5/8)*(5/8) = 100/576
ADDED= 1200/5184+200/576 =125/216
125+91=216 great.
I draw a probability tree and listed all the probabilities on it.
I ended up with 9 possible scenarios and 4 of them were favourable.
If I work out the probabilities of each of the favourable ones and add them together then hopefully that will be the answer.
Although I would check it by adding all possible outcomes probabilities together (They should add to 1)
