How do you do this? on Probability(without replacement)

[ak]

A bag contains 4 yellow balls and "n" red balls. Two balls are drawn without replacement. Which expression represents the probability that one ball is yellow and ball is red?

P.S. the answer is (4/n+4)(n/n+3) + (n/n+4)(4/n+3)

 

[chroot]

First, what's the probability of drawing a yellow ball? There are four yellow balls, out of (n+4) total balls, so the probability of drawing a yellow ball is:

$$\frac{4}{n+4}$$

Now, after having drawn a yellow ball, there are only three yellow balls and n red balls out of a total of n+3 balls left. Since there are n red balls and n+3 total balls, the probability of drawing a red ball is:

$$\frac{n}{n+3}$$

Now, the probability of the first event occurring (drawing a yellow ball), followed by the second event occurring (drawing a red ball) is the product of the two probabilities. In other words, when you have two events that occur in succession, you multiply their probabilities together. When you multiply them together, you get:

$$\frac{4}{n+4} \cdot \frac{n}{n+3}$$

That's the probability for drawing a yellow ball, then drawing a red ball.

Of course, there are two ways to draw one yellow ball and one red ball -- the way we've already described, yellow followed by red, and another way, red followed by yellow. That case, red followed by yellow, is represented by the second half of the expression you listed:

$$\frac{n}{n+4}\frac{4}{n+3}$$

When there are more than one way to get a result (one red ball and one yellow ball), the probabilities of each way add together. Thus, the probability of getting one red ball and one yellow ball is the sum of the probability of drawing yellow then red and the probability of drawing red then yellow.

- Warren

 

[tacman]

To find the probability of one yellow ball and one red ball being drawn without replacement, we can use the formula P(A and B) = P(A) * P(B|A), where P(A) is the probability of event A occurring and P(B|A) is the probability of event B occurring after event A has already occurred.

In this case, event A is drawing a yellow ball and event B is drawing a red ball. So, the probability of drawing a yellow ball first is 4/n+4, since there are 4 yellow balls out of a total of n+4 balls in the bag. Then, the probability of drawing a red ball after the yellow ball has been removed is (n/n+3), since there are n red balls left out of a total of n+3 balls.

So, the overall probability of drawing one yellow ball and one red ball without replacement can be expressed as (4/n+4)(n/n+3), as well as (n/n+4)(4/n+3). These two expressions are equivalent and can be simplified to (4n)/(n+4)(n+3), which is the answer provided.