Question 1153647: Hi, I got stuck in a question:There are 5 Black balls, 4 red balls, and 1 blue ball in a bag. If 3 balls are drawn without replacement, find probability of:
a) Exactly 1 black ball is drawn given that at least 1 black ball is drawn.
For a), I have tried to use conditional probability, but I can't reach the value smaller than 1,and so here I am seeking for help, looking forward to your reply and thanks for your effort.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! probability of 1 black ball is 3*(5/10)(5/9)(4/8)=300/720=50/120, because 3 ways to choose
that is (5C1)(5C2)/10C3=5*10/120=50/120 or 5/12
for 2 it would be 3*5C2*5C1, which is the same
for 3 it would be 5C3/10C3=10/120, or 1/12
Given that at least 1 (11/12 probability from adding the above)
this would be (5/12)/(11/12)=5/11
to check, probability of no black balls being drawn is 5C0*5C3/10C3, which is 1/12
Answer by ikleyn(52754)  (Show Source):
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