1.An airport shuttle bus makes 4 scheduled stops for 15 passengers. What is the probability that all of them get off at the same stop and also the probability that atleast one gets off at each stop.
- You are told that of the four cards face down on the table two are red and two are black. If you guess all four at random, what is the probability that you get 0,2,and 4 right.
pretty easy probability questions i fumbled on the first one though due to nervousness.
Simple combinatorics;
total number of ways in which 15 people go into four different doors is simply a multi-set with (n+k-1)C(K) = (4+15-1)C(15). Now if all 15 go into one door, there are four different possibilities. So the probability that all get off in one stop is 4/18C15. For Part 2, the denominator is simply the same which 18C15. For the numerator we use the same theory; but now assuming 1 gets of at each stop, we have 11 choices and we use the same multi-set theorem and we see that it is; (4+11-1)C(11) which is simply 14C11. So the answer is 14C11/18C15.
I admittedly know nothing about multi set combinatorics, so please indulge my forthcoming ignorance.
Why is the solution to the first question not 4 * (1/4)^15 ? Each man has a probability of 1/4 of going into any particular door, so all fifteen going into the first door would be (1/4) ^ 15, and then multiply by the 4 doors.
Thanks
indian-banker is incorrect. The odds that all 15 go into one door is different from the probability that 14 go into one door, while 1 goes into another door. Hence you can't do the n choose k.
For part ii), take 1 - answer to part (i)
The second question requires clarification: at random? What are your choices? (just red and black?)
Damn I still have interviews coming up, and don't remember any statistics/probability. Any recommendations on how to brush up on these sort of questions? Thx.
I am guessing that this was for an S&T interview?
Yes it was. This was the first time i had a interviewer bring in a book and read off probability questions to me. It was really strange, but good experience.
For the second question:
0 correct = 4 correct = 1/2 ^ 4 = 1/16
2 correct = 1/2 ^ 4 * 4C2 = 6/16 = 3/8
4 correct should be 1/6. The question should be clarified. My interpretation of "random guess" should be based on given the knowledge of 2 black cards and 2 red cards. Which is to say, you can't guess 4 black, or 3 red and 1 black.
otherwise, it should just be a ordinary binomial distribution.
I assumed you knew it was 2 red cards and 2 black cards, too. If this is the case, it would be impossible to get 1 or 3 cards correctly guessed. The answer would then 100%.
0 correct = 4 correct = 1 / 4c2 = 1/6 1 correct = 3 correct = 0 / 4c2 = 0 2 correct = 4 / 4c2 = 2/3
Which firm?
silver94 you are wrong. you don't choose NCK, if you read my answer you choose (n+k-1)C(k). It's a multi-set. Read "a walk through combinatorics" by bona, and you will see similar questions. Think about it this way. There are 15 people that need to be allocated into 4 different doors. So that's the denominator or total number of choices. There are four doors, so the answer is 4 divided by 18C15. For the second part, you can't simply do 1-part 1. I can't even begin to tell you how wrong that answer is. 1 - part 1 essentially says the probability that not all 15 people go into the same door. As for Lemur's comment, I'm actually not great at probability so I really don't know why your answer would be wrong, but I do know that these types of questions are typically answered easily with combinatorics. I will think about why (1/4)^15*4 doesn't work and get back to you.
Here is another way to think about it: You have 15 people and 3 possible separators. The people to the left of the 1st separator go to the 1st stop, people between 1st and 2nd separators get off at the 2nd stop, between 2nd and 3rd separators get off at the 3rd stop and to the right of the 4th separator get off at the 4th stop.
Let A1 - A15 be the passengers, and || be the separators. Then they will all be arranged in some kind of order like A1 - A5 || A6 - A10 || A11 - A13 || A14 - A15
So we have 15 + 3 = 18 total objects that have to be organized, and we have to pick 3 locations for the 3 separators so there are 18C3 possible combinations of stops. But there are only 4 combinations where all the passengers are either
1)To the left of the 1st separator 2)Between the 1st and 2nd 3)Between the 2nd and 3rd 4)To the right of the 3rd separator
So our answer is 4/18C3 = 4/18C15 as indian-banker said
Thanks, that explanation makes sense intuitively. Do you know why the logic behind 4 * (1/4)^15 as the answer is flawed?
The answer is not flawed. 4 * (1/4)^15 is the probability of that event. You are considering 15 ppl are distinct.
The combinatorial way should be 4 / [(4+15-1)C(15)]. It is the number of specific events over all possible events. You don't consider ppl are distinct. so order doesn't matter.
Out of interest BudFox24, what degree do you study?
Indian banker is right and i studied economics with a minor in math and philosophy
I believe it has to do with the people being indistinguishable, correct me if I am wrong.
If the people are labeled Person 1, Person 2, Person 3, etc than there would be 4^15 different combinations. However, in this case the people are indistinguishable, therefore this overestimates the amount of combos.
!
This is absolutely correct! Cheers!
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