Question 1

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?

Accepted Answer

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.

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.

Question 2

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?

Accepted Answer

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.

Question 3

why is this question not a combination because we can shuffle the order in which these people are chosen for the positions?

Accepted Answer

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!

Question 4

What is the difference between conditional probability and dependent probability?

Accepted Answer

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).

Question 5

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!

Accepted Answer

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

Question 6

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?

Accepted Answer

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.

Question 7

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? :'(

Accepted Answer

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.

Question 8

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!

Accepted Answer

In permutations, the order matters.  In combinations, it doesn't.

Question 9

why cant you use combination for this question?

Accepted Answer

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

9_P_3 = 9!/6! = 9 * 8 * 7 = 504

Question 10

Is this considered a permutation in a sense since it did need a specific order?

Accepted Answer

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.

Statistics and probability

Course: Statistics and probability > Unit 8

Example: Different ways to pick officers

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