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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)
I do this one and leave the other as excersises.Quote:
Originally Posted by valo4212
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%
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?