conditional probability homework problem :-(

[lilwolf]

A box contains one yellow , two red, and three green balls. Two
balls are randomly chosen without replacement. Define the following
events:
A:{one of the balls is yellow}
B:{at least one ball is red}
C:{both balls are green}
D{both balls are the same color}

P(B|A')=
P(D|B')=
P(C|A)=0

P(A'|B)=
P(B'|D)=
P(D|C)=

i knew one anyway *s*


i`m not sure how to start this really usually i make a list
or tree or venn diagram , or can just look and eliminate
to do these , usually the problems will have numbers or i have
to fill in the diagram, but none have had balls and colors
i`m not sure if i need to use the probability like 1/6 for the
yellow ball and try and figure it out or even if the sample
set will be all the balls or will be different on each one.
The formulas we have been using are like
P(A|B)=P(A u B)/ P(B)
P (A u B)= P(A)x P(A|B)
P (A n B)= P(A) x P(B)
P (A|B)= P(A)
P (B|A)= P (B)
etc.

 

[galactus]

A box contains one yellow , two red, and three green balls. Two
balls are randomly chosen without replacement. Define the following
events:

Forget about those formulas. They'll just confuse you.

I will show you these to show you how you can approach them.
Perhaps then, you stand a better chance of working similar problems.

A:{one of the balls is yellow}

If the first drawn is yellow and the second draw is any of the other 5: \(\displaystyle \L\\\frac{1}{6}\cdot\frac{5}{5}\)

If the first is any of the other 5 and the second is a yellow:
\(\displaystyle \L\\\frac{5}{6}\cdot\frac{1}{5}\)

Add them.

Or you can just find the prob. of drawing yellow and multiply by 2.

B:{at least one ball is red}

The opposite of at least one is none. Find the probability of none are red and subtract from 1. \(\displaystyle \L\\1-\frac{4}{6}\cdot\frac{3}{5}\)

C:{both balls are green}

Probability second is green given first is green. Think it through.
The prob. of the first being green is 3/6. Since there's no replacement, you have 5 balls left of which 2 are green. So, the prob. of the second green is 2/5.
\(\displaystyle \L\\\frac{3}{6}\cdot\frac{2}{5}\)

D{both balls are the same color}

Can't be yellow in that event, since there's only one of them. So, what's the prob. of both red or both green. \(\displaystyle \L\\\underbrace{\frac{3}{6}\cdot\frac{2}{5}}_{\text{both green}}+\overbrace{\frac{2}{6}\cdot\frac{1}{5}}^{\text{both red}}\)

P(B|A')=
P(D|B')=
P(C|A)=0

P(A'|B)=
P(B'|D)=
P(D|C)=

i knew one anyway *s*


i`m not sure how to start this really usually i make a list
or tree or venn diagram , or can just look and eliminate
to do these , usually the problems will have numbers or i have
to fill in the diagram, but none have had balls and colors
i`m not sure if i need to use the probability like 1/6 for the
yellow ball and try and figure it out or even if the sample
set will be all the balls or will be different on each one.
The formulas we have been using are like
P(A|B)=P(A u B)/ P(B)
P (A u B)= P(A)x P(A|B)
P (A n B)= P(A) x P(B)
P (A|B)= P(A)
P (B|A)= P (B)
etc.[/quote]