# Math Help - Probability question

1. ## Probability question

I'm just having some difficulty working out how to begin with regards to the following question - "A group of dogs managed to escape from an enclosure. Find the smallest number of dogs in the group if the probability that at least one of the dogs in the group has less than four spots is greater than 0.99."

Probability of less than four spots is = 0.7

Any help would be greatly appreciated, thanks.
- desitter

2. ## Re: Probability question

Hey deSitter.

Your question doesn't seem to make sense: what does the number of spots have to do with the number of dogs that escape?

3. ## Re: Probability question

I think the idea is to find some way to link the number of dogs to the probability of their spots, and then relate that to the 0.99 probability, then solve for the approximate number of dogs that would equal that probability (rounded to the nearest integer). My initial idea was to use a binomial distribution to sum all the probabilities of having at least one dog with 4 spots, equating this to 0.99 and solving for n. The problem is that without knowing an interval this can't be done.

4. ## Re: Probability question

I'm also trying to get my head around a conditional probability question that does not quite feel intuitive to me. The question is - "Find the probability that a person saw film type 'A' the first time if they see a film type 'B' the second time." I'm used to conditional probability questions being the other way around, with the condition being an event that preceded the probability in question. Below is a tree diagram I have drawn with the probabilities of each event. The first choice of film type is 50/50 while there is a 92% chance of seeing film type 'B' given that they previously saw film type 'A', and a 95% chance of seeing film type 'B' having previously seen film type 'B'.

My inclination is to use the conditional probability formula, but I'm unsure how to write it given the the wording of the question. Pr(Type 'A' as the first film | They see film type 'B' as the second film). What is the interception between the probability we are after and the condition? I tend to try and visualize these conditional probabilities on tree diagrams, usually these questions just point to a child branch or series of child branches after a parent node that is the condition [e.g. Pr(b2 | b1) = 0.95], but this one is difficult for me to visualize. What probability exactly is the question after? Can it be drawn as a branch on the tree diagram I have drawn? If so can someone point this out to me?

- desitter

5. ## Re: Probability question

I'd like to take stab at this, if you don't mind.

$\\\text{Let }A_{F}\text{ be the event that a person saw Film A first and let }B_{F} \text{ be the event that a person saw Film B first.}$

$\\\text{Let }A_{S}\text{ be the event that a person saw Film A second and let } B_{S} \text{ be the event that a person saw Film B second.}$

So, we know

$Pr[A_{F}]=Pr[B_{F}]=0.5$

$Pr[B_{S}|A_{F}]=0.92$

$Pr[B_{S}|B_{F}]=0.95$

We're looking for

$Pr[A_{F}|B_{S}]=\frac{Pr[A_{F}\cap B_{S}]}{Pr[B_{S}]}$

From what we're given, we know

$Pr[B_{S}\cap A_{F}]=[B_{S}|A_{F}]*P[A_{F}]=0.92*0.5=0.46$

$Pr[B_{S}\cap B_{F}]=[B_{S}|B_{F}]*P[B_{F}]=0.95*0.5=0.475$

$Pr[B_{S}]=Pr[B_{S}\cap A_{F}]+Pr[{B_{S}\cap B_{F}]=0.46+0.475=0.935$

Now we have everything we need

$Pr[A_{F}|B_{S}]=\frac{Pr[A_{F}\cap B_{S}]}{Pr[B_{S}]}=\frac{0.46}{0.935}=0.49197861$

It turns out that your tree diagram was right on track. You didn't even need my 2 cents.

6. ## Re: Probability question

For the first question you posted, you're on the right track using the binomial distribution and solving for n.

You're given

$Pr[X\geq1]=0.99$

But remember that

$Pr[X\geq1]=1-Pr[X<1]=1-P[X=0]$

That should make it much easier to solve for n

7. ## Re: Probability question

Great stuff downthesun01, I initially found the answer to question 2 just as you did but was not 100% convince it was correct; the working seemed right but it felt very unintuitive. I'm still a bit unsure what this probability represents (in a tree diagram / graphical sense). For example, I can easily see how the conditional probability formula works from something like Pr(b2 | b1), as Pr(b2 intersect b1) = b2 * b1, therefore b2 * b1 / b1 would just be b2. In problem 2 I can't step through the working in the same way. Perhaps you could help me understand how this problem works in similar terms? Regardless though your '2 cents' confirms my initial answer, which boosts my confidence quite a bit, so it's certainly appreciated.

I jotted down your hint to my first problem before making the connection to the binomial distribution, therefore neglecting to consider it. Doh! So thank you for the reminder. I arrived at a value for 'n' of approximately 3.82498, hence the minimum number of dogs required should be 4 for a probability of > 0.99. An alternative question to this one that I am now working on is similar to the previous, accept for it being 'at least eight dogs' rather than 'at least one'. At the moment I can only think to have Pr(X ≥ 8) = 1 - sum binomial distributions form n-choose-0 to n-choose-7, but this solution, although probably correct, would require more working than I think the question requires. There must be a simpler way, any suggestions?

8. ## Re: Probability question

Another quick question, for Markov chains, how would I go about calculating the expected value for a particular number of iterations? Say I have an initial state matrix 's' and a transition matrix 't', the expected value of t^4*s = ?. I'm thinking I should sum each of the probabilities multiplied by the numerical value assigned to each state, e.g. 1*0.08 + 2*0.0524 + 3*0.0515 + 4*0.0515, just like you would for discreet random variables, but I'm not all that convinced this is correct. It's not like each step value is a value that needs weighting. Perhaps just a sum of the probabilities divided by 4?

9. ## Re: Probability question

If you had a matrix of values (and assuming you don't want the long term equilibrium distribution), you basically need to specify a prior for the starting value (i.e. the 0th transition) and use that to get your expectation.

So an as an example if we had three states then we have a prior distribution with three probabilities P(X0 = a), P(X0 = b), P(X0 = c) and using the fact P(A and B) = P(A|B)P(B) we use this probability to get the expected values.

More specifically we have E[Z] = Summation for all values of z and x P(Z = z, X0 = x) where P(Z = z, X0 = x) = P(Z = z|X0 = x)*P(X0 = x).

If you aren't given specifics you can assume P(X0 = x) is a discrete uniform.

10. ## Re: Probability question

Originally Posted by deSitter
I'm just having some difficulty working out how to begin with regards to the following question - "A group of dogs managed to escape from an enclosure. Find the smallest number of dogs in the group if the probability that at least one of the dogs in the group has less than four spots is greater than 0.99."
One possibility is that exactly one dog escaped and it has less than four spots!

Probability of less than four spots is = 0.7
Where did this come from? Is it an additional problem? It must not be part of the original problem because it is outside the quotes.

Any help would be greatly appreciated, thanks.
- desitter

11. ## Re: Probability question

Hi HallsofIvy, the probability 0.7 comes from summing up a few probabilities given in a discrete distribution from a previous question. The probabilities are given as 2 = 0.3, 3 = 0.4, 4 = 0.23, 5 = 0.05 and 6 = 0.02. Attached is my working for the question if you would like to look it over.

@Chiro, I'm afraid I am not quite able to following your working. My knowledge is not particularly great. Things are not explained particularly well in the class I am in (distance education), my weekly work is comprised of being directed to read a textbook chapter and submit some questions, and I rarely get any opportunity to ask questions. The textbooks are notorious for lacking any substantial explanation, and rather just focus on memorization of formulas, which makes it quite an uphill battle. If I give you a practical example, could you step me through the theory?

Transition matrix = T
[ 0.95 0.92 ]
[ 0.05 0.08 ]

Initial state matrix = s
[ 0 ] <--- initial probability of seeing film B
[ 1 ] <--- initial probability of seeing film A

What would be the expected number of times of seeing either film, being that four films were seen after the initial film A (T^4*s)?

12. ## Re: Probability question

Anybody have any thoughts on the above problem?

Thank you,
- deSitter