Each box contains a pen with one of four colors, equally likely. Each box is independent.

After opening N (N=4) boxes, what is probability you have a full set (all 4 colors)?

My daughter came up with the following solution which I verified with a quick Matlab run.

P(set)=P(any color in box 1)*P(different color in box 2)*P(different color in box 3)*P(different color in box 4)

P(set)=1*.75*.5*.25

P(set)=.094

I tried a more formal method (because the next problem requires a higher value of N) and got a different answer.

P(set)=P(at least one blue)*P(at least one red|at least one blue)*P(at least one yellow|at least one red & blue)*P(at least one green|at least one red & blue & yellow)

P(at least one blue)=[1-P(no blue in any box)]=[1-.75^4]=.684

P(at least one red|at least one blue)=[1-P(no red in 3 boxes)]=[1-.75^3]=.578

P(at least one yellow|at least one red & blue)=[1-P(no yellow in 2 boxes)]=[1-.75^2]=.438

P(at least one green|at least one red & blue & yellow)=[1-P(no green in last box)]=[1-.75]=.25

So, P(set)=.684*.578*.438*.25=.043

I know the first answer is right. I don't know why the second method doesn't work. Can someone please point out my mistake? ]]>

There are tickets for a documentary film and exactly people to buy the tickets (one person can buy only one ticket).

There are ticketbooths for selling the tickets and each ticketbooth will sell exactly tickets.

Suppose the researcher labels the people with ID according to their arrivals.

Say, the first arrival buys his/her ticket from one of the ticketbooths.

Then, the second arrival buys his/her ticket from one of the ticketbooths. The second arrival can buy his/her ticket from the same ticket booth that the first arrival had bought or from one of the other two ticketbooths.

In this way, the last arrival buys his/her ticket.

The researcher has recorded from which ticketbooth which arrivals have bought the tickets.

Suppose from ticketbooth A, arrival #3, #5, #6, #12 have bought the tickets.

From ticketbooth B, arrival #1, #2, #9, #11 have bought the tickets.

From ticketbooth C, arrival #4, #7, #8, #10 have bought the tickets.

In how many ways the people can buy tickets from the ticketbooths?

#My Attempt:

If inside a ticketbooth the order of the ID doen't matter, then the number of ways the people can buy tickets from the ticketbooths is

But for my example, since inside a ticketbooth it is naturally ordered (that is, first arrival comes before second arrival and so on), should I consider the order inside a ticketbooth?

For the above example, is ORDER more meaningful or NOT? ]]>

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I cant get my head round the following exercise:

You are testing dice for a casino to make sure that sixes do not come up more frequently than expected. Because you do not want to manually roll dice all day, you design a machine to roll a die repeatedly and record the number of sixes that come face up. In order to do a Bayesian analysis to test the hypothesis that p = 1/6 versus p = .175 , you set the machine to roll the die 6000 times. When you come back at the end of the day, you discover to your horror that the machine was unable to count higher than 999. The machine says that 999 sixes occurred. Given a prior probability of 0.8 placed on the hypothesis p = 1/6 , what is the posterior probability that the die is fair, given the censored data? Hint - to find the probability that at least x sixes occurred in N trials with proportion p (which is the likelihood in this problem), use the R command :

1-pbinom(x-1,N,p)

The possible answers are 0.5, 0.684, 0.8 or 0.881.

I would really appreciate if someone could help me here! I need to understand this approach!

Cheers and thanks in advance! Markus ]]>