Question 1177405: A box contains 4 white balls and 3 red balls. Three balls are drawn from the box without replacement. What is the probability of drawing:
a) 3 white balls?
b) 3 red balls?
c) 2 white balls?
d) What is the expected number red balls?
e) What is the expected number of white balls?
thank you
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 7C3 ways of drawing the balls=35 possible outcomes
a. 4C3*3C0 numerator=4, so 4/35 ways to getting 3 white balls.
b. this is 3C3*4C0=1, so 1/35 ways to get 3 red balls.
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look at 1 white ball 4C1*3C2=12
look at 2 white balls 4C2*3C1=18
Those possibilities for the numerator should add up to 35. They do
E(x)=x*p(x)
for 0: 0*1/35=0
for 1: 1*(12/35)=12/35
for 2: 2*(18/35)=36/35
for 3 3*(4/35)=12/35
E(W)=60/35 for white balls
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E(both) should be 3, so it ought to be 45/35 E(R), which it is.
3 red balls 1/35 probability=3/35
You don't compute the probability of 4 balls being chosen because the problem limits it to 3.
2 red balls is 12/35 probability =24/35
1 red ball is 18/35 probability=18/35
0 red balls is 4/35 probability=0 ; yes, this was an error.
The question was raised about the probability of drawing 4 white balls. It is 0, because the question refers to only three balls being drawn, not four. The expected value of white balls drawn, WHEN THREE BALLS ARE DRAWN, is what is given above.
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