A bucket contains orange tennis balls and yellow tennis balls from which 5 balls are selected at random, but assume that the bucket contains 7 orange balls and 6 yellow balls.

(1) What is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow?

(2) What is the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow?

zwdefrgthyjujyhbgvfvgbhyjuhgvfgtbhynbfcdfvrgbthvfcdxdcfrvgtbfrrgtbhyn

(1) From 1, subtract the probability that none are yellow and the probability that none are orange. Every other possibility has one or more of both.

All orange probability = (7/13)(6/12)(5/11)(4/10)(3/9) = 0.0163
All yellow probability = (6/13)(5/12)(4/11)(3/10)(2/9) = 0.0047
1-0.0163-0.0047 = 0.979

(2) See what you can do using a similiar method.

Thanks! I am still working on the second part.. and having a difficult time... but I am sure I will eventually figure it out. :)

From one, subtract the probabilities of:

(1) zero or one orange, (2) zero or two yellow and (3) one orange and one yellow.
Zero orange (all yellow):0.0047
Zero yellow (all orange):0.0163
One orange (4 yellow)): 5x(6/13)(5/12)(4/11)(3/10)(7/9)= 0.08159
One yellow (4 orange): 5*(7/13)(6/12)(5/11)(4/10)(6/9)= 0.1632
1-0.0047-0.0163-0.0816-0.1632 = 0.8974

To find the probabilities of selecting balls from a bucket, we need to use the concept of combinations.

Let's solve each question step by step:

(1) What is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow?

To find the probability that at least one ball is orange and at least one ball is yellow, we need to find two separate probabilities: the probability of selecting only orange balls and the probability of selecting only yellow balls. We can then subtract these probabilities from 1 to get the final result.

Now, let's find the probability of selecting only orange balls:
There are 7 orange balls in the bucket, and we want to select all 5 balls as orange. This can be represented as selecting a combination of 5 orange balls out of 7. The formula for combinations is:
C(n, r) = n! / (r! * (n-r)!)
where n is the total number of items, and r is the number of items to be selected.

In this case, we want to select 5 balls out of 7 orange balls, so the combination is:
C(7, 5) = 7! / (5! * (7-5)!) = 21

The total number of combinations to select 5 balls from the total pool of 13 balls (7 orange + 6 yellow) is:
C(13, 5) = 13! / (5! * (13-5)!) = 1,287

Therefore, the probability of selecting only orange balls is 21/1,287.

Similarly, the probability of selecting only yellow balls can be calculated as:
C(6, 5) / C(13, 5) = 6/1,287.

Now, we subtract these probabilities from 1 to get the final result:
1 - (21/1,287) - (6/1,287).

Therefore, the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow is approximately 0.985.

(2) What is the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow?

To find this probability, we need to calculate the probability of selecting two orange balls and three yellow balls, as well as the probability of selecting three orange balls and two yellow balls. We can then add these two probabilities together to get the final result.

The probability of selecting two orange balls and three yellow balls can be calculated as:
C(7, 2) * C(6, 3) / C(13, 5).

The probability of selecting three orange balls and two yellow balls can be calculated as:
C(7, 3) * C(6, 2) / C(13, 5).

Adding these two probabilities will give us the final result.

Therefore, the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow can be calculated by summing the probabilities of selecting two orange and three yellow balls, and selecting three orange and two yellow balls.