High school statistics
- Probability using combinations
- Example: Lottery probability
- Example: Different ways to pick officers
- Probability with permutations & combinations example: taste testing
- Probability with combinations example: choosing groups
- Probability with combinations example: choosing cards
- Probability with permutations and combinations
- Mega millions jackpot probability
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?(43 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.(18 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.(10 votes)
- What is the difference between conditional probability and dependent probability?(4 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)
- 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.(11 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!(4 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)
- Is this considered a permutation in a sense since it did need a specific order?(2 votes)
- 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)
- 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)
- in how many ways 1 or 9 person enter in a room(2 votes)
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.