A group of teachers are required to prepare 4 questions each for a weekly quiz. The number of questions made in a week has a Poisson distribution with a mean of 6.
- Find the probability that the group manages to write enough questions for the upcoming weekly quiz.
My understanding:
Let X be denoted as the number of questions in a week.
Poisson distribution with $\lambda = 6$: $X \sim Po(6)$
We are to find $P(X \geq 4)$ which is $0.848796117223352$ (calculated).
- As some teachers in the group are busy with other responsibilities, there is a probability of 40% that only half of the group will work on the quiz questions. If only half of the group work on the questions then only 3 questions are prepared on average. If the group fails to finish 4 questions that week, what is the probability that only half the group worked on the questions?
My understanding:
I know that we are to find $P(50\%\text{ of team})$ if $X < 4$ but I'm not sure how to incorporate the probability of 40%. Perhaps my understanding is incomplete.
- The principal decides to increase the number of questions in the weekly quiz and asks the group of teachers to use every question created. If a student has a probability of 40% in answering the questions correctly and is expected to answer 2 questions correctly in the upcoming quiz, what is the probability that the group of teachers worked on creating the questions that week?
As you can see, I was only only able to answer question 1 and would need some help in the remainder 2 questions. This is a new topic to me and any help is greatly appreciated!
(2)
Let $A$ is a random variable denote that only half of the group will work on the quiz questions.
So we know that P(A) = 0.4, and A is bernoulli distribution with parameter p=0.4.
And we also know that given A, the conditional distribution $X|A$ is poisson(3).
The problem is "given $X<4$, the probability that $A$ is true". So we want to find out $P(A|X<4)$. By the Bayesian theorem:
$P(A|X<4)=\frac{P(A,X<4)}{P(X<4)}=\frac{P(X<4|A)P(A)}{P(X<4)}=\frac{P(X<4|A)P(A)}{P(X<4|A)+P(X<4|A^c)}$
where $A^c$ is "more than half of the group will work on the quiz questions".
We have already known $P(X<4|A)$ because $X|A$ is Poi(3). $P(A)$ is 0.4 because of bernoulli, and $P(X<4|A^c)$ is Poi(6). So now you can find the probability of $P(A|X<4)$.
(3)
I'm not sure if I understand this question well...
A student has probability 0.4 in answering the questions correctly and is expected to answer 2 questions correctly. We can assume that the number of answering questions correctly is Binomial(n, p=0.4), so the expected correctly answer $\mu=np=0.4n=2$, so $n=5$.
Thus the question become to "given that the number of questions is 5, find the probability of $A^c$", that is $P(A^c|X=5)$.
Again using Bayesian theorem:
$P(A^c|X=5)=\frac{P(X=5|A^c)P(A^c)}{P(X=5)}=\frac{P(X=5|A^c)P(A^c)}{P(X=5|A^c)+P(X=5|A)}$
We have already known $X|A^c$ is Poi(6), $X|A$ is Poi(3), so the probability can be easily computed.