2014 Extension 1 BOS Trial Exam Discussion Thread (1 Viewer)

[mattjd]

I was wondering whether someone could please give a hint for 13) part b) ii) and iii).
Initially I thought to find P(k+1)/P(k) = 1/((n-1)(k+1)(m-k+1)) and find the smallest k for which P(k+1)/P(k) < 0. If k = m+1, then P(k+1)/P(k) = 0 and if k = m + 2, P(k+1)/P(k) = 1/((n-1)(m+2)(-1) < 0 so presumably k = m + 2 is the most likely number of times the chosen number will appear. Which isn't what the question says.

Then I tried calculating the expected value which is a different interpretation of the question which might have worked
E(k) is the expected value of the random variable k, where E(k) = sum from k = 0 to k= m of kP(k)
So I calculated E(k) using the substitution k(m choose k) = m((m-1) choose (k-1)), made the substitution u = k-1, and expanded the binomial (1 + (n-1))^(m-1) to simplify the expression to get E(k) = m/n which is closer to what the given answer was, but still not right.

 

[Trebla]

I was wondering whether someone could please give a hint for 13) part b) ii) and iii).
Initially I thought to find P(k+1)/P(k) = 1/((n-1)(k+1)(m-k+1)) and find the smallest k for which P(k+1)/P(k) < 0. If k = m+1, then P(k+1)/P(k) = 0 and if k = m + 2, P(k+1)/P(k) = 1/((n-1)(m+2)(-1) < 0 so presumably k = m + 2 is the most likely number of times the chosen number will appear. Which isn't what the question says.

Then I tried calculating the expected value which is a different interpretation of the question which might have worked
E(k) is the expected value of the random variable k, where E(k) = sum from k = 0 to k= m of kP(k)
So I calculated E(k) using the substitution k(m choose k) = m((m-1) choose (k-1)), made the substitution u = k-1, and expanded the binomial (1 + (n-1))^(m-1) to simplify the expression to get E(k) = m/n which is closer to what the given answer was, but still not right.

I'm not exactly sure why you want to solve for the smallest k such that P(k+1)/P(k) < 0 when no such k exists given that P(k) and P(k+1) are probabilities and the total number of trials is m (so the largest possible value of k is m).

The approach for (ii) is to find the largest value of k such that P(k+1) > P(k).

 

[mattjd]

Hahaha, thankyou just realised that what I should have done was P(k+1)/P(k) > 1, which is then equivalent to P(k+1) > P(k).

 

[jmoyo]

official solutions?

 

[RealiseNothing]

I was wondering whether someone could please give a hint for 13) part b) ii) and iii).
Initially I thought to find P(k+1)/P(k) = 1/((n-1)(k+1)(m-k+1)) and find the smallest k for which P(k+1)/P(k) < 0. If k = m+1, then P(k+1)/P(k) = 0 and if k = m + 2, P(k+1)/P(k) = 1/((n-1)(m+2)(-1) < 0 so presumably k = m + 2 is the most likely number of times the chosen number will appear. Which isn't what the question says.

uhhh how can probability be negative?

 

[taylorswiftxoxo]

where are solutions???

 

[deboiz]

HAHAH sorry to dig this up again but does anyone know where i could get solutions?

 

[Carrotsticks]

HAHAH sorry to dig this up again but does anyone know where i could get solutions?

https://www.dropbox.com/s/42h5l5pes5a1hpr/Extension 1 BOS Trial 2014 Solutions & Criteria.pdf?dl=0

 

[braintic]

https://www.dropbox.com/s/42h5l5pes5a1hpr/Extension 1 BOS Trial 2014 Solutions & Criteria.pdf?dl=0

Either I am misinterpreting the question, or the answer to question 14a is not correct.

If n=2, k=2, you are making ORDERED arrangements of two distinct digits, so the answer should be 10P2.

But your formula gives 10C2.

(The question talks of 'combinations', yet it says that order is important, so it's not clear what you are asking for.)