What is the probability of getting two red balls from box II?

[Monoxdifly]

There are two boxes. Box I contains 5 red balls and 4 white balls. Box II contains 3 red balls and 6 white balls. A ball is taken randomly from box I and put into box II. Then, from box II, two balls are taken randomly. What is the probability of both balls taken from box II are red?

I only know that the probability of getting red ball from box I is $$\frac{1}{9}$$. How to do the rest?

 

[MarkFL]

I'm thinking we need to find the probability that a red ball is selected from box I AND then 2 red ball are selected from box II OR a white ball is selected from box I AND then 2 red balls are selected from box II. Let's look at the first event. The probability that a red ball is selected from box I is equal to ratio of the number of red balls to the toal number of balls:

$$P(\text{red ball from box I})=\frac{5}{9}$$

Okay, now in box II we have 4 red ball and 6 white balls. So, the probability of drawing 2 red balls is:

$$P(\text{2 red balls from box II})=\frac{4}{10}\cdot\frac{3}{9}=\frac{2}{15}$$

And so the probability of both sub-events happening is:

$$P(\text{red ball from box I AND 2 red balls from box II})=\frac{5}{9}\cdot\frac{2}{15}=\frac{2}{27}$$

Can you compute the probability that a white ball is selected from box I AND then 2 red balls are selected from box II?

 

[Monoxdifly]

$$\frac{4}{9}\cdot\frac{3}{10}\cdot\frac{2}{9}$$

 

[MarkFL]

Okay, after we reduce we have:

$$P(\text{white ball from box I AND 2 red balls from box II})=\frac{4}{135}$$

And recall we found:

$$P(\text{red ball from box I AND 2 red balls from box II})=\frac{2}{27}$$

So, what do you suppose we should do with these two probabilities?

 

[Monoxdifly]

Add?

 

[MarkFL]

Add?

Yes, with "OR" we add...so what do you get?

 

[physiclawsrule]

In case you've taken a white from the 1st box - 5/9 x (3/10 x 2/9)
In case you've taken a red from the 1st box - 5/9 x (4/10 x 3/9)

 

[Monoxdifly]

Yes, with "OR" we add...so what do you get?

$$\frac{14}{135}$$

 

[MarkFL]

$$\frac{14}{135}$$

Yes, that's what I got as well. :D