Ball Probability

[mcarvell]

Hi All,

I have a question about the probability of two events

The question I have been given is as follows

1. There are 3 blue, 1 white and 4 red identical balls inside a bag. If it is aimed to take two balls out of the bag consecutively, what is the probability to have 1 blue and 1 white ball?

a. 3/28
b. 1/12
c. 1/7
d. 3/7

I keep coming to the probability of 1 blue which is 3/8 * 1 white which would be 1/7 if the blue ball is not replaced 1/8 if it is. Either way it is 3/56 non replacement or 3/64 replacement. Any idea what I am doing wrong here? Thanks for any help!

Matt

 

[pka]

I have a question about the probability of two events
The question I have been given is as follows
1. There are 3 blue, 1 white and 4 red identical balls inside a bag. If it is aimed to take two balls out of the bag consecutively, what is the probability to have 1 blue and 1 white ball?
a. 3/28
b. 1/12
c. 1/7
d. 3/7
I keep coming to the probability of 1 blue which is 3/8 * 1 white which would be 1/7 if the blue ball is not replaced 1/8 if it is. Either way it is 3/56 non replacement or 3/64 replacement. Any idea what I am doing wrong here?

Use combinations: \(\displaystyle \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{8}{2}}=~?\)

 

[tkhunny]

Without Replacement

Draw #1 - Blue 3/8
----- Draw #2 - White 1/7 - Success
----- Draw #2 - Other 6/7 - Failure
Draw #1 - White 1/8
----- Draw #2 - Blue 3/7 - Success
----- Draw #2 - Other 4/7 - Failure
Draw #1 - Other 4/8 - Failure

Checking Draw #1 3/8 + 1/8 + 4/8 = 8/8 = 1 - We got it all.

3/8 * 1/7 + 1/8 * 3/7 = 3/56 + 3/56 = 6/56 = 3/28

Just pay better attention. Let notation help you. Stay organized and make sure you are complete and thorough.

 

[mcarvell]

Thanks guys! math isn't my greatest subject so that was very helpful