Asked by kd

A bag contains 3 black balls, 4 green balls and 5 yellow balls.
a) if two balls are picked at random without replacement, find the probability that Botha balls are of the same color.
b) How many black balls must be added to the bag so that the probability of drawing a black ball is 1/2.

Answers

Answered by oobleck
P(black,black) = 3/12 * 2/11
do the same steps for P(green,green) and P(red,red)
then add them all up, since the only ways to get two of the same color is to get two blacks, greens, or reds.

There are 12 balls in all
There are 3 black balls
so, you need to add x black balls to make them 50% --
12+x = 2(3+x)
Answered by Reiny
a) you got 3 cases: both black, both green, or both yellow
without replacement, prob of both black = (3/12)(2/11) = 1/22
without replacement, prob of both green = .....
without replacement, prob of both yellow = ...
add them up

b) number of blacks to be added ---- x
So now you have x+3 black balls out of 12+x in total
prob (black) = (x+3)/(x+12) = 1/2
solve for x
Answered by R_Scott
ways of picking two balls ... 12C2 ... 66
... two black balls ... 3C2 ... 3
... two green balls ... 4C2 ... 6
... two yellow balls ... 5C2 ... 10

p(both same color) = (3 + 6 + 10) / 66

b) (3 + x) / (12 + x) = 1/2
Answered by Tumusiime
thank you
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