Welcome to a second video on probability. The goal of this video is to determine
the probability of different events. This video builds and what we learned from
the introductory video on probability. And we'll also be using combinations to
determine sum of the probabilities. So just to review, the probability of an
event compares the favorable number of outcomes to the total number of outcomes.
And this can be expressed as a fraction, decimal, or percentage.
Let's go and take a look at some examples. Here we're using a standard
deck of cards, and for number one want to determine the probability of drawing
two cards that are both face cards
Well we're selecting two cards from the deck, so the total number of ways to select
two cards from a deck would be fifty-two, choose two. And the total number of ways of selecting
two face cards, well there are twelve face cards in the deck, so that would be twelve,
choose two. Remember we can write this using different notation.
For example, we can write twelve, choose two over fifty-two, choose two.
And I say that because we're going to go ahead and use the calculator to evaluate this,
and the calculator uses this type of notation.
Let's go ahead and evaluate this probability. So we have twelve, choose two. So twelve, math,
probability, option three. And then we type in r, which is two.
That's sixty-six, and we'll divide this by fifty-two, choose two.
So fifty-two, math, probability, option three, and two, enter.
So it's approximately zero, point, four, nine, eight.
Now I do want to take a look at this problem in a second way.
We don't have to use combinations. We could think of this
as two events, where the first event is selecting the first card, and the second
event is selecting the second card. And we can ask ourselves, what is the
probability of selecting a face card as the first card.
Well there's twelve face cards out of fifty-two cards,
time's the probability of selecting a face card a second time, given this is
already a face card.
Well now there would only be fifty-one cards and there would also only be eleven face cards left.
So we'd have eleven over fifty-one for the second probability. And this product will give
us the same probability. Let's go ahead and check it.
So we'll have twelve divided by fifty-two, times eleven divided by fifty-one.
And as you see, the result is the same. So many times there's more than one way
to find a given probability. So this is also a valid way to solve this probability question.
Number two, the probability of drawing two cards that are both aces.
Okay, so with the probability of two aces. Well again we're selecting two cards, so
the total number of ways of selecting two cards is still going to be fifty-two, choose two.
But there's only four aces in a deck, so the favorable number of ways of
selecting two aces would be, there's four aces and we want to choose two a time.
And let's go ahead and use the calculator again, to do this.
So we have four, choose two
divided by fifty-two, choose two.
And we have approximately point, zero, zero, four, five.
It should make sense that this probability is much less then number one,
because here we have twelve face cards, and here we only have four aces.
And again the alternative approach would be to think of this as
two different events, so we could multiply the probabilities together
where the first probability is the probability of selecting an ace. On the
first card, which would be four out of fifty-two times, three out of fifty-one.
And again this product will result in the same probability.
Let's go and take a look at number three.
The probability of drawing four cards all of the same suit.
Okay, so this one gets a little more interesting. The probability of selecting
four cards of the same suit.
Let's first consider the denominator. Well we're selecting four cards now so the
denominator would be fifty-two, choose four. Remember in all of these we're using
combinations, because the order doesn't matter. If the order didn't matter we'd be
using permutations. Now in the numerator we want to have four cards of the same suit.
Well each suit contains thirteen cards, so part of the favorable number of ways would be
thirteen, choose four. However, there are four different suits to choose from. So we
would have. So this would actually be four times thirteen, choose four. Again remember
the first card could be any of the four suits
therefore, we have four times as many ways to have a favorable outcome.
Let's go ahead and determine what this would be, and we'll take a look at this one a different way as well.
The numerator would be four times thirteen, choose four, and we're going to divide this by
fifty-two, choose four.
This comes out to approximately zero, one, zero, six. Again, let's take a look at
an alternative approach. We're drawing four cards. One, two, three, four.
And let's consider each card individually.
This first card can actually be any card
as long as the other three cards are of the same suit.
So the probability of selecting the first card correctly would be fifty-two out of fifty-two.
However, for the second card, whatever suit this is, this one must match it.
So there's one less card. And whatever suit this is there's going to be one less of that
suit to select from. So there would be twelve instead of thirteen left in that suit.
And then from there, the total number of cards drops by one, and so does the number of cards
in that suit. This would be eleven over fifty. This would be forty-nine.
And this would drop from eleven to ten.
Let's go and make sure this product is the same as what we have over here
with the combinations.
Well the first fraction is one. Then we have twelve divided by fifty-one.
And then we have eleven divided by fifty. And ten divided by forty-nine.
And you can see the result is the same.
So this just emphasizes you often have more than one way to find a given
probability. Let's see if we have time for some more questions.
A group of ten students made up of six females, and four males form a
committee of four. What is the probability the committee is all male? Well the total
number of outcomes would be the number of ways we can select a committee of four
from a group of ten that would be ten, choose four.
The favorable ways of selecting an all male committee, well there's four males, so we'd have four, choose four
in our numerator. Let's go to the calculator. Well four, choose four is equal to one,
so we'll have one divided by ten, choose four.
So the probability is approximately point, zero, zero, four, eight, to have a
committee of all males. Number two, what is the probability the community will
be all female? Well it's a similar process, so we have the probability of all
females. We're still selecting four from a group of ten, so our denominator is ten, choose
four. Now the favorable number of ways of selecting females, since there are six
females, we have six choose, four. And let's see what this would be. We have six choose
four divided by ten, choose four. And you can see it is higher, because there are more females to select
from. It's approximately point, zero, seven, one, four.
Okay on this question here it says, what is the probability the comittee is made up of
two females and two males? Well our denominator is still going to be ten,
choose four.
The number of ways of selecting two females, if there's six to
select from that would be six, choose two, times a number of ways to select two
males. And there are four males to select from, so that would be four, choose two.
So the total number of ways of selecting the two females times the total number of
ways of selecting the male's, divided by the total number of ways to select the
entire committee, would give us the desired probability.
Let's go back to the calculator. Here our numerator would be six, choose two, times four, choose two.
So there's our numerator and we'll divide by ten, choose four.
And there we have it. Approximately point, four, two, eight, six.
Okay that's going to do it for this video.
I hope you found the explanations helpful.