# Probability of selecting the first urn.

• Aug 8th 2006, 08:23 AM
kwtolley
Probability of selecting the first urn.
Ok here it goes,
One urn contains 3 red and 7 white marbles. A second urn contains 2 red and 8 white marbles. One urn is selected and a ball drawn. The probability of selecting the first urn is 0.4 and the probability of selecting the second is 0.6. If a white ball is drawn, find the probability it came from the second urn.

3 red 7 white, urn 1= 10
2 red 8 white, urn 2= 10
P(urn_1)=0.4, P(urn_2)=0.6 0.4*0.6=24
looks like to me urn 2 has a better chance than 1 becaue of th 60%, both have 10 marbles but urn 2 has one more white one than urn 1. List of possible answers are -

1. 0.800
2. 0.632
3. 0.560
4. 0.480 From here I'm stuck in the mud. Need help with the rest of it. thanks for looking over my problem.
• Aug 8th 2006, 08:41 AM
kwtolley
man that was easy
I tryed everything I could to come up with an answer, I didn't think about that % is all that was needed. Thanks soooo much for showing me that. I have a few more like this I have to do, they all will work out like that. Thanks again Galactus for all your help.
• Aug 8th 2006, 09:54 AM
Soroban
Hello, kwtolley!

I hope you're familiar with Bayes' Theorem . . .

Quote:

Urn A contains 3 red and 7 white marbles.
Urn B contains 2 red and 8 white marbles.
One urn is selected and a ball drawn.
The probability of selecting urn A is 0.4
and the probability of selecting urn B is 0.6.

If a white ball is drawn, find the probability it came from the urn B.

$\displaystyle 1)\;0.800\qquad2)\;0.632\qquad3)\;0.560\qquad4)\;0 .480$

Bayes' Theorem: .$\displaystyle P(\text{urn B }|\text{ white}) \;= \;\frac{P(\text{urn B }\land\text{ white})} {P(\text{white})}$

The numerator is:
. . $\displaystyle P(\text{urn B }\land\text{ white})\:=\:P(\text{urn B})\cdot P(\text{white})$ $\displaystyle \:=\:(0.6)(0.8)\:=\:0.48$

The denominator is:
. . $\displaystyle P(\text{white}) \:=\:P(\text{urn A }\land\text{ white}) + P(\text{urn B }\land\text{ white})$
. . . . . . . . .$\displaystyle =\qquad\quad(0.4)(0.7) \qquad+ \qquad(0.6)(0.8)\quad =\quad0.28$$\displaystyle + 0.48 \;= \;0.76$

Therefore: .$\displaystyle P(\text{urn B }|\text{white}) \;= \;\frac{0.48}{0.76} \;= \;\frac{12}{19} \;=\;0.631578947$

Answer (2): .$\displaystyle 0.632$

• Aug 8th 2006, 10:45 AM
galactus
Soroban, I sure had a brain fart :o

I even ran Bayes theorem and thought "naaaaa". I checked back to fix my stupidity and you had taken care of it. Good!. Sorry kwtolley
• Aug 8th 2006, 08:59 PM
kwtolley
thanks and yes
Yes I have read Bayes Theorem but I need to see it on paper how to set up a problem the first time. Now I see it and I have some more to do. I can say one thing it's not as easy as it looks if you ask me.