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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.

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  • blobby green style avatar for user Adam
    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?
    (43 votes)
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    • spunky sam blue style avatar for user Vince
      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.
      (18 votes)
  • blobby green style avatar for user Dawn Cannon
    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)
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    • spunky sam blue style avatar for user Vince
      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.
      (10 votes)
  • blobby green style avatar for user rajiv.abraham
    What is the difference between conditional probability and dependent probability?
    (4 votes)
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    • marcimus pink style avatar for user Priyanka Kumar
      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)
  • male robot johnny style avatar for user Hoolala
    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)
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  • blobby green style avatar for user DawoodJabbarQureshi
    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!
    (4 votes)
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    • orange juice squid orange style avatar for user John Paul Marceaux
      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)
  • piceratops ultimate style avatar for user AtMetaphase
    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)
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    • leafers sapling style avatar for user Peter Collingridge
      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)
  • purple pi purple style avatar for user zella
    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)
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  • leaf green style avatar for user Daniel Mielnik
    Is this considered a permutation in a sense since it did need a specific order?
    (2 votes)
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  • blobby green style avatar for user Adnan Younis
    why cant you use combination for this question?
    (2 votes)
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  • blobby green style avatar for user Rajan Savaliya
    in how many ways 1 or 9 person enter in a room
    (2 votes)
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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.