Question

Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. What is the probablility of two or three of the balls being white.

Answer 1

\[P(2\ white)=\frac{3C2\times5C2}{8C4}=you\ can\ calculate\]
\[P(3\ white)=\frac{5C1}{8C4}=you\ can\ calculate\]
The probability of 2 or 3 of the balls being white is given by:
P(2 white) + P(3 white) = you can calculate.

Answer 2

I've never seen it written like that, so I'm not sure what that means...How would it look if I was to plug that into a calculator?

Answer 3

(3!\2!)(5!/2!)/8!/
4!4!?

Answer 4

The equations show combinations. For example 3C2 means the number of combinations of 3 things taken 2 at a time.
\[\frac{\frac{3!}{2!}\frac{5!}{2!3!}}{\frac{8!}{4!4!}}\]

Answer 5

do you know where I could put that into to get the answer, my calculator doesn't like this problem

Answer 6

Does your calculator have a nCr function?

Answer 7

Yes

Answer 8

Try entering 3, then select nCr (you might have to press a function or shift key first), finally enter 2 and press =.
What do you get?

Answer 9

3

Answer 10

So now can you do the calculations?

Answer 11

3/112?

Answer 12

\[\frac{3C2\times5C2}{8C4}=\frac{3\times10}{70}\]
\[\frac{5C1}{8C4}=\frac{5}{70}\]

Answer 13

The probability of 2 or 3 of the balls being white is given by: P(2 white) + P(3 white) =
\[\frac{30}{70}+\frac{5}{70}=0.5\]