Question

Sampling without replacement: An urn contains 2 black and 3 white balls. 2 balls are drawn at random without replacement. The probability that the sample contains at most one black ball.
Supposedly the answer is .90, but I am not sure how to get that.

Answer 1

oooh AT MOST

Answer 2

compute the probability that there are no black balls
compute the probability that there is one black ball
then add

Answer 3

Probability No Black
\[
\frac 3 5 \frac 2 4=\frac 3 {10}
\]
Probability One Black
\[
\frac{2\ 3}{5\ 4}+\frac{3\ 2}{5\ 4}=\frac 3 5
\]
Add them you obtain .9

Answer 4

@satellite73 the tree that I made is



P( no black): P(WW)= (3/5)(2/4) --> .3
P( 1 back): P(BW) + P(WB)= (2/5)(2/4) + (3/5)(2/4) --> .5

.3 + .5= 0.80 not 0.90