A cooler contains 7 cans of cola: 6 regular colas and 4 diet colas. If 3 cans of cola are selected at random, what is the probability that 2 cans of regular cola and 1 can of diet cola are selected?

Sorry, I meant to say 10 cans NOT 7.

That would equal the probability of getting one diet cola only, on the first, second or third selection.

3 * (4/10)*(6/9)*(5/8) = 1/2

To find the probability, we first need to calculate the total number of possible outcomes and the number of favorable outcomes.

Total number of ways to select 3 cans out of 7 cans:
This can be calculated using the combination formula: nCr = n! / (r!(n-r)!)
In this case, n (total number of cans) = 7 and r (number of cans selected) = 3
So, 7C3 = 7! / (3!(7-3)!) = 7! / (3!4!) = (7*6*5) / (3*2*1) = 35

Number of ways to select 2 regular colas out of 6 regular colas:
This can be calculated using the combination formula again.
6C2 = 6! / (2!(6-2)!) = 6! / (2!4!) = (6*5) / (2*1) = 15

Number of ways to select 1 diet cola out of 4 diet colas:
4C1 = 4

Now, we can calculate the probability by taking the favorable outcomes divided by total outcomes.
Probability = (Number of favorable outcomes) / (Total number of outcomes)
Probability = (Number of ways to select 2 regular colas) * (Number of ways to select 1 diet cola) / (Total number of ways to select 3 cans)
Probability = (15 * 4) / 35 = 60 / 35 ≈ 1.71

Therefore, the probability of selecting 2 regular colas and 1 diet cola is approximately 1.71.

To find the probability, we need to determine the total number of possible outcomes and the number of favorable outcomes.

First, let's find the total number of possible outcomes. We are selecting 3 cans from a cooler that contains 7 cans of cola. Therefore, the total number of possible outcomes is given by the combination formula:

nCr = n! / (r!(n-r)!)

where n is the total number of items, and r is the number of items selected.

For this problem, n = 7 (total number of cans) and r = 3 (number of cans selected). Plugging in the values, we have:

7C3 = 7! / (3!(7-3)!)
= 7! / (3!4!)

Now let's calculate the number of favorable outcomes. We want to select 2 cans of regular cola and 1 can of diet cola. We have 6 regular colas and 4 diet colas in the cooler. Using the same combination formula, we have:

6C2 = 6! / (2!(6-2)!)
= 6! / (2!4!)

4C1 = 4! / (1!(4-1)!)
= 4! / (1!3!)

Since we want both cans of regular cola and the can of diet cola, we multiply these two values:

P(2 regular colas and 1 diet cola) = (6C2) * (4C1)

Finally, we can calculate the probability:

Probability = Favorable outcomes / Total outcomes
= (6C2) * (4C1) / (7C3)

Simplifying, we get:

Probability = (15) * (4) / (35)
= 60 / 35
= 12 / 7

Therefore, the probability of selecting 2 cans of regular cola and 1 can of diet cola is 12/7 or approximately 0.8571.