00:01
All right, so this problem says we have a group of 12 kids, sorry, 12 students, where seven of them are females and five of them are males, and they want to form a committee out of those 12 of five.
00:14
Okay.
00:16
The first question that asks us is what is the probability if they make that committee and choose the kids randomly that all five of those students will end up being the five males? okay, so the way we're going to do that is we're going to look at each choice individually.
00:32
He would say for the first student that's chosen, there's a 5 out of 12 chance that it would be one of the guys.
00:41
The next student, it would be a 4 out of 11 chance, right? because it can't choose the same kid again.
00:47
So there's four males left out of the 11 remaining students.
00:50
It would be 4 out of 11.
00:51
So the next one would be 3 out of 10.
00:53
Next one would be 2 out of 9.
00:55
And then for the last guy, he would be 1 out of only 8 students left.
00:59
So that would be his probability.
01:02
Okay.
01:02
What we do is we multiply all of those probabilities together to find the total probability of that, of all of them being chosen occurring.
01:13
Okay, so we're going to do this by multiplying all of the tops of the fraction together.
01:18
So we would take five times four, then it's three, tens two, and times one, because that is 120.
01:30
Okay, and then we're going to multiply all of the bottom part of it together as well.
01:33
So 12, sorry about the dog, times 11, times 10, times 9, times 8.
01:52
Okay, and then define the probability, we just divide those two.
01:55
So it's 120 divided by 95 -040.
02:03
It comes out as that.
02:05
0 .00 -126 would be our probability there.
02:13
Or if we wanted to write it as a fraction, we could simplify this, right? can we divide this by 10? this becomes 12 over 9504.
02:39
We can divide both of those by 2.
02:45
It can be divided by 2 again.
02:50
So now we have 3 over 2376.
03:02
And these are both divisible by 3.
03:07
Comes out the probability of 1 in 792.
03:13
So either of these two answers here would be correct for part a.
03:18
Let me draw a better box around these.
03:27
All right, for part b, it wants to know what is the probability that the committee is at most to females.
03:33
Okay, so the way we're going to do that, anytime you see something that says at most or at least, we're going to look at each of the individual possibilities.
03:44
And then add them all together, right? so if we're looking at a committee that at most has two girls on it, what we would say is we need to know the probability of there being one girl, sorry, one female and four males.
04:04
We need to know the probability of there being two females and three males, and the probability of there being zero females and five males.
04:15
Those are our three options, right? for there to be at most two girls.
04:20
We already know this one.
04:21
That's what we found up here.
04:23
It's 0 .00126.
04:27
Okay, for two females and three males, the order doesn't matter.
04:31
So we can do it in any order, but let's say we pick the guys first, just because you already got it kind of figured out up here.
04:37
Right? we would say five over 12 for the first guy, then four over 11, then three over 10, because there are three guys.
04:48
And then for the two girls.
04:49
Girls that are left, it would be seven out of the nine that are left over, right? and then six out of eight after that.
05:04
Okay, so we would multiply all of those together.
05:08
Five times four times three times seven.
05:16
Okay, it comes two thousand five hundred twenty over the same value because these are all the same as they were before, right? so two thousand five hundred 120 over 95 -040.
05:31
Actually, let's do these interactions.
05:34
It'll make our lives easier in the end.
05:37
So this would be 120 over that.
05:45
This one is 2 ,520 over that.
05:51
And then this one's the same idea, except for instead of three guys, we have four.
05:56
So we would replace these two at the end here with another guy, so two out of the nine remaining guys or people...