### Instruction: The Grammar club has 20 members: 10 boys and 10 girls.  A 4-person committee is chosen at random.  What is the probability that the committee has at least 1 boy and at least 1 girl?### Response: I want to find the probability of the event that the committee has at least 1 boy and at least 1 girl.
This event is the complement of the event that the committee has only boys or only girls.
So, I can use the formula for the probability of the complement of an event: P(A) = 1 - P(A').
To find P(A'), I need to find the probability of the event that the committee has only boys or only girls.
This event can happen in two ways: either the committee has 4 boys or 4 girls.
These two cases are mutually exclusive, meaning they cannot happen at the same time.
So, I can use the formula for the probability of the union of two mutually exclusive events: P(A or B) = P(A) + P(B).
To find P(A), the probability of the event that the committee has 4 boys, I need to count how many ways I can choose 4 boys out of 10 boys, and divide that by the total number of ways I can choose 4 people out of 20 people.
This is a combination problem, because the order of the people in the committee does not matter.
I can use the formula for the number of combinations: nCr = n!/(r!(n-r)!), where n is the total number of items, and r is the number of items chosen.
So, the number of ways I can choose 4 boys out of 10 boys is 10C4 = 10!/(4!(10-4)!) = 210.
And the total number of ways I can choose 4 people out of 20 people is 20C4 = 20!/(4!(20-4)!) = 4845.
So, P(A) = 210/4845.
Similarly, to find P(B), the probability of the event that the committee has 4 girls, I need to count how many ways I can choose 4 girls out of 10 girls, and divide that by the total number of ways I can choose 4 people out of 20 people.
This is also a combination problem, and I can use the same formula.
So, the number of ways I can choose 4 girls out of 10 girls is 10C4 = 10!/(4!(10-4)!) = 210.
And the total number of ways I can choose 4 people out of 20 people is still 20C4 = 20!/(4!(20-4)!) = 4845.
So, P(B) = 210/4845.
Now, I can find P(A') by adding P(A) and P(B).
P(A') = P(A) + P(B) = 210/4845 + 210/4845 = 420/4845.
Finally, I can find P(A) by subtracting P(A') from 1.
P(A) = 1 - P(A') = 1 - 420/4845 = 4425/4845.