Bag A contains \(3\) white and \(2\) red balls. Bag B contains \(6\) white and \(3\) red balls. One of the two bags will be chosen at random, and then two balls will be drawn from that bag at random without replacement. What is the probability that the two balls drawn will be the same color? Express your answer as a common fraction.

https://latex.artofproblemsolving.com/b/8/f/b8f02d08a1f352c38211990a099609f56b24d017.png

tertre Feb 20, 2018

#2**+2 **

OK...here is MY guesstimate.....I think people are just waiting for ME to humiliate myself with a wild answer !

Bag One how many ways can you withdraw 2 of the balls out of 5 c 2 = 10 ways

how many ways to pair two whites 3 c 2 = 3

How many ways to pair two reds 2 c 2 = 1

Summary bag one (3+1)/10 and we have a 1/2 chance of grabbing bag 1

1/2 (4/10) = 4/20

Bag 2 how many ways to withdraw two out of 9 balls 9 c 2 = 36 ways

how many ways to pair two whites 6 c 2 = 15

how many ways to pair 2 reds 3 c 2 = 3

Summary bag 2 (15 +3)/36 = 18/36 and we have a 1/2 chance of grabbing bag 2

1/2 (18/36) = 18/72

Now let's add them together: 4/20 + 18/72 = 648/1440 = ** 9/20**

**(Well...right or wrong...that is what I derived !) ~EP**

ElectricPavlov Feb 21, 2018

#3**+1 **

Your solution looks good, EP.

**Condensed Form**

\(\dfrac{1}{2} * \large (\Large \rho \normalsize \text{(two balls of same color bag A)} + \Large \rho \normalsize \text{(two balls of same color bag B)}\large)\\ \text { }\\ \large \dfrac{1}{2} \small \dfrac{\dbinom{3}{2}+\dbinom{2}{2}}{\dbinom{5}{2}} +\large \dfrac{1}{2} \small \dfrac {\dbinom{6}{2}+\dbinom{3}{2} }{\dbinom{9}{2}} =\Large \dfrac{9}{20} = \ 0.45 \)

GA

GingerAle Feb 21, 2018