# Probability help with marble!

• Mar 7th 2007, 04:42 PM
valo4212
Probability help with marble!
A box contains 3 red marbles, 6 blue marbles, and 1 white marble. The marbles are selected at random, one at a time, and are not replaced. Find the probability.
P(blue and red)
P(white and blue)
P(red and white)
P(red and white and blue)
P(red and red and blue)
P(red and blue and blue)
P(red and red and red)
P(white and blue and blue)
P(white and red and red)
• Mar 7th 2007, 06:49 PM
ThePerfectHacker
Quote:

Originally Posted by valo4212
A box contains 3 red marbles, 6 blue marbles, and 1 white marble. The marbles are selected at random, one at a time, and are not replaced. Find the probability.
P(red and white and blue)

I do this one and leave the other as excersises.

How many ways are there of choosing 3?
In total we have 3+6+1=10.
Thus, 10C3=120

How many ways of choosing 1 white, 1C1=1
Similarly, red: 3C1=3 and blue: 6C1=6.
Thus, there are (1)(3)(6)=18 different ways of choosing red,white and blue.

The probability therefore is,
18/120=3/20=15%
• Mar 8th 2007, 05:36 AM
Soroban
Hello, valo4212!

Quote:

A box contains 3 red marbles, 6 blue marbles, and 1 white marble.
The marbles are selected at random, one at a time, and are not replaced.
Find the probabilities.

From the wording of the problems,
. . I assume that the colors are drawn in that order.

. . . . . . . . . . . . . . . 6 . .3 . . . . .1
(1) .P(B and R) . = . --- · --- . = . --
. . . . . . - - - . . . . . 10 . .9 . . . . .5

. . . . - - - . . . . . . . . 1 . . 6 . . . . 1
(2) .P(W and B) . = . --- . -- . = . ---
. . . . . . - - - . . . . . .10 . .9 . . . .15

Want to try the rest of them now?