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### Course: Statistics and probability > Unit 8

Lesson 4: Combinatorics and probability- Probability using combinations
- Probability & combinations (2 of 2)
- Example: Different ways to pick officers
- Example: Combinatorics and probability
- Getting exactly two heads (combinatorics)
- Exactly three heads in five flips
- Generalizing with binomial coefficients (bit advanced)
- Example: Lottery probability
- Probability with permutations and combinations
- Conditional probability and combinations
- Mega millions jackpot probability
- Birthday probability problem

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# Example: Different ways to pick officers

Thinking about the different ways we can pick officers in order to find the probability of one situation in particular. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Shouldn't the answer be 9*8*7/3! instead of 9*8*7 because it says that the officers are chosen at random and so the order does not matter?(48 votes)
- No. You're using the combination formula. That's incorrect because order DOES matter here; it's a permutation problem. There's a difference between Marsha, Sabita, Robert and Robert, Marsha, Sabita because the order is associated with the role. You can order the roles in whatever way you want (VP, Secretary, President), but the order that you pick has an impact.

If the question was, "There are 9 candidates for 3 board positions. What's the probability that Marsha, Sabita and Robert will be on the board," then you would be right because order wouldn't matter then.(41 votes)

- Im not sure where this falls in probability. But I have been trying to figure out how to set this problem up to solve it. Can you please help?? The question is "If the probability of surviving a Head on car crash at 55 MPH id 0.056, then what is the probability of not surviving."How do I set this up to solve it I am so confused?(5 votes)
- There are only 2 options: survive and not survive. If you add up the probabilities of all possible options, then they must equal 1 (or 100%). In this case, if the probability of surviving is .056, then the probability of not surviving is 1 - .056, which is .944.(12 votes)

- I do understand how to solve this question using combinations, but I don't get why this method doesn't work. "There are 7 students in the class with 2 boys and 5 girls. If 4 of them were picked at random, what is the probability that all of them will be girls?" Using combinations, I would have (4C5) / (7C4) and get 1/7. However, why doesn't this work? Since there are 5 girls to start with out of 9 students, the probability that the first student picked will be a girl is 5/9. Then, the probability that the second student picked will be a girl would be 4/8. The third will be 3/7 and the fourth, 2/6. if I multiply those 4 fractions, the product will be 5/126, which is incorrect. I don't get why the second method doesn't work.... Can you please help me? :'((2 votes)
- There are only 7 students, not 9. The first fraction should be 5/7. the next three are 4/6, 3/5, and 2/4. When multiplied together the result is 1/7. This method does work as long as you do not make any mistakes.(12 votes)

- why is this question not a combination because we can shuffle the order in which these people are chosen for the positions?(4 votes)
- Right, but if we shuffle the positions, will Sabita always get vice president for example? Here the order matters because the order goes from: President - Vice president - Secretary. So if I have 3 people A,B,C then if we say something like ABC, that would mean person A is president, and B is VIP. If we get BCA, then the roles would change once again. Hope this helped!(6 votes)

- What is the difference between conditional probability and dependent probability?(3 votes)
- I think both of them are the same. Conditional probability is the one in which outcome of the second event depends upon the outcome of the first event. Take for an example, we have two events A and B. If we have performed the event A ( taking sample space A union B).Now its the turn for B to happen, then we will be considering sample space of A to find the probability of B( i.e.Probability of B in A).(7 votes)

- A little question someone hopefully can help me with, I've had a similiar task, however this is how I thought It would be solved, : P = president, VP = vice president, S = secretary

9/P * 8/VP * 7/S + 9/VP * 8/S * 7/P + 9/P * 8/VP * 9/S *8/P * 7/VP = 504 + 504 + 504 = 1512 possibilities, I know its wrong, but iv used this method somewhere else in a probabily task, in which case would my anser be right?

Thanks again!(3 votes)- I believe you may have mistypes in your question. You wrote that the terms in the first statement would equal 504+504+504. However, your third term (9/P * 8/VP * 9/S *8/P * 7/VP) would not equal 504 it would equal 36288. But I am assuming that this is the typo, and the equation you meant to type was : (9/P * 8/VP * 7/S) + (9/VP * 8/S * 7/P) + (9/P * 8/VP * 7/S) = 504 + 504 + 504 = 1512

Addition is used in probability to count events that are independent rather than dependent on one another. The question posed by sal is an example of a dependent event because one change in any position (say Marsha was chosen for vice president instead of president) changes the entire event into something totally different. Under our question M/P S/VP R/S and S/P M/VP R/S are two entirely different events, thus the events are dependent.

Now your question includes addition and adds the term that we counted for the one event. This would be present in a question that asked something to the effect of: "What is probability of Marsha for president, Sabita for Vice President, Marshal for Secretary; AND two other events that have similar structure to this event like two other clubs with 9 members choosing three positions as well. In this case these events would be independent because the other clubs' positions choices would not affect the positions of the club we are discussing in the problem. Independent events can normally be spotted with the word AND(6 votes)

- I can't seem to figure out when you find the total number of possibly outcomes by multiplying X options by number of events ( heads or tails = 2 options, 4 = number of events so 2*2*2*2) versus when you would use the number of events factorial as total possible events (4! = number of events). Can anyone clarify this for me?(3 votes)
- If you flip a coin and get heads, that doesn't mean that you can't get heads on the second flip. In other words, events can be repeated. So for each flip there are two possible outcomes, so the number of possibilities for four flips is 2*2*2*2 = 16.

In the example in the video, we are picking people for positions in a society, but each person can only have one position. In other words, we can't repeat an event (where the event is picking someone). This is the same as picking balls from a bag**without**replacing them. So after each event, there is one less choice for the following event. That's why we use a factorial. If, in the example in the video, each person could hold all three titles, then instead, the probability would be 1 in 9 * 9 * 9.(4 votes)

- This may have already been asked, but can someone please explain the difference between permutations and combinations? And a clear definition for each? Thanks so much!(2 votes)
- In permutations, the order matters. In combinations, it doesn't.(5 votes)

- why cant you use combination for this question?(2 votes)
- Since order matters in the case as we want a specific person to be in each officer position, we could use a permutation here and in effect that is what he did, even if he didn't state that's what he was doing.

9_P_3 = 9!/6! = 9 * 8 * 7 = 504(4 votes)

- Is this considered a permutation in a sense since it did need a specific order?(1 vote)
- Hey Daniel!

Yep, your right.

You can view it as a permutation since the order matters in which the peoples are chosen (the President is not the same as a Vice President).

If it would not be of interest who will be chosen first or second or third, it would be a combination.(5 votes)

## Video transcript

A club of nine people wants
to choose a board of three officers, President, Vice
President, and Secretary. Assuming the officers are chosen
at random, what is the probability that the officers
are Marcia for president, Sabita for Vice President,
and Robert for Secretary? So to think about the
probability of Marcia-- so let me write this-- President is
equal to Marcia, or Vice President is equal
to Sabita, and Secretary is equal to Robert. This, right here, is one
possible outcome, one specific outcome. So it's one outcome out of the
total number of outcomes, over the total number of
possibilities. Now what is the total number
of possibilities? Well to think about that, let's
just think about the three positions. You have President, you
have Vice President, and you have Secretary. Now let's just assume that we're
going to fill the slot of President first. We don't
have to do President first, but we're just going
to pick here. So if we're just picking
President first, we haven't assigned anyone to any officers
just yet, so we have nine people to choose from. So there are nine possibilities
here. Now, when we go to selecting our
Vice President, we would have already assigned one
person to the President. So we only have eight
people to pick from. And when we assign our
Secretary, we would've already assigned our President and Vice
President, so we're only going to have seven people
to pick from. So the total permutations here
or the total number of possibilities, or the total
number of ways, to pick President, Vice President, and
Secretary from nine people, is going to be 9 times 8 times 7. Which is, let's see,
9 times 8 is 72. 72 times 7, 2 times 7 is 14, 7
times 7 is 49 plus 1 is 50. So there's 504 possibilities. So to answer the question, the
probability of Marcia being President, Sabita being Vice
President, and Robert being Secretary is 1 over the total
number of possibilities, which is 1 over 504. That's the probability.