Probability tree

Question

Use a probability tree to find the probabilities of the indicated outcomes.

An urn contains 7 red, 8 white, and 9 black balls. One ball is drawn from the urn, it is not replaced, and a second ball is drawn. (Enter your probabilities as fractions.)

(a) What is the probability that both balls are white?

(b) What is the probability that one ball is white and one is red?

(c) What is the probability that at least one ball is black?

Apparently all my answers were wrong :(

Answer 1

a)

P(2W) = 8/24 x 7/23

= 1/3 x 7/23

= 7/69= probability both balls are white

when there are 8 white balls and 2 balls selected randomly without replacement, given 24 total balls (7+8+9=24)

on the 1st draw, 1/3 of the balls are white

on the 2nd draw 7/23 of the balls are white

multiply those two fractions to get the probability of both white

8/24 x (8-1)/(24-1)

= 1/3 x 7/23 = 7(1)/3(23) = 7/69

c)

Probability at least one ball is black = 1 minus probability no balls are black

P(zero Black) = (24-9)/24 x (23-9)/23

= 15/24 x 14/23

= 5/8 x 14/23

=5(14)/8(23)

= 35/92

P(at least one Black) = 1-35/92

= 92/92 -35/92

= (92-35)/92

= 57/92

b)

Probability of one white and one red

= P(WR) + P(RW)

= probability of 1st white, 2nd red

+ probability 1st red, 2nd white

= 8/24 x 7/23 + 7/24 x 8/23

= 1/3 x 7/23 + 7(8)/24(23)

= 1/3 x 7/23 + 1/3 x 7/23

= 2/3 x 7/23

= 14/69

not sure why this problem is listed under the category "calculus" It fits far better under "probability" or "statistics"

Maybe you should tell us the answers you tried that didn't work? we could see better where you might have gone wrote. Unfortunately, the above answers don't include "trees" which seem to be required

Sometimes you can't see the forest for the trees

or if you had the appropriate computer program, do a computer simulation

Answer 2

You have a total of 24 balls in the urn, it looks like the ball is never replaced, so we will be selecting from 24 balls during our first selection, and 23 balls during our second.

A) There are 8 white balls from the start, after selecting a white ball, 7 remain. Thus the probability that both are white would be (8/24)(7/23) = (7/69)= .1014

B) Using the same logic, (8/24)(7/23) (selecting white then red) OR (7/24)(8/23) (selecting red then white). Either way, you get the same probability. (56/552) = (7/69)= .1014

C) The probability of selecting a black ball is (9/24). Now we have at least one black ball, our second drawing does not matter (it could be black or any other color). So our answer is (9/24)= (3/8) = .375