Thread: Probability of selecting the first urn.

1. 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.

2. 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.

3. Hello, kwtolley!

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

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): .$\displaystyle 0.632$

4. Soroban, I sure had a brain fart 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

5. 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.

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