Suppose that 28 crayons, of which four are red, are divided random among Jack, Marty, Sharon, and Martha (seven each). If Sharon has exactly one red crayon, what is the probability that Marty has the remaining 3?
No idea...workin on the same problem.
0.0263
To find the probability that Marty has the remaining 3 red crayons given that Sharon has exactly one red crayon, we can use conditional probability.
First, let's calculate the total number of ways the crayons can be divided among Jack, Marty, Sharon, and Martha. We have a total of 28 crayons, and we want to divide them equally among the four people, so each person receives 7 crayons.
The total number of ways to distribute the crayons can be calculated using the concept of combinations. The formula for combinations is:
nCr = n! / (r! * (n-r)!)
In this case, we want to distribute 28 crayons equally among 4 people, which means we need to calculate the number of ways to choose 7 crayons for Jack, 7 crayons for Marty, 7 crayons for Sharon, and 7 crayons for Martha.
So, the total number of ways to distribute the crayons equally among the four people is:
Total ways = 28C7 * 21C7 * 14C7
Next, let's calculate the number of favorable outcomes where Marty has the remaining 3 red crayons while Sharon has exactly one red crayon.
Since there are 4 red crayons in total, and Sharon has one of them, there are 3 red crayons remaining. We need to distribute these 3 red crayons among the other 3 people (Jack, Marty, and Martha).
The number of ways to distribute the 3 red crayons equally among the 3 people is:
Favorable ways = 3C3 * 21C7 * 14C7
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of outcomes:
Probability = Favorable ways / Total ways
So, to find the probability that Marty has the remaining 3 red crayons given that Sharon has exactly one red crayon, you would calculate:
Probability = (3C3 * 21C7 * 14C7) / (28C7 * 21C7 * 14C7)