Extra exercise during class

Mister_JWO · 2022-10-19 00:22:24

During class our math teacher gave us an extra exercise to make this week. We can use every source we want to find the solution. Can someone help me with the work method to solve this problem?

A man is participating in a television program. He has to follow a way and on 4 points he has to do a task. When he fails a task he has to go back to point 1. On the first point he has 85% chance to succeed, on the second point 65%, on the third point 45% and on the fourth point he has 25% chance to succeed. When he has passed the fourth point he wins. How much task has the man in average to do before he wins?

I’ve put it already into a graph to make it more visual:
https://drive.google.com/file/d/1IGCPFTl3buc2TALFFmz3pyBrHHXPHsTB/view?usp=sharing

The solution doesn't have to be a whole number.

Bob · 2022-10-19 04:54:55

hi Mister_JWO

Welcome to the forum.

I think I've got a method for this. LATER EDIT. I am worried I may have oversimplified so I'm going to run a simulation program.

Let's start with the probability that he succeeds first go with all four tasks. Please work out the probability for this; call it p.

If he doesn't that must happen with probability (1-p)

So the probability that he succeeds second go is (1-p).p

and third go (1-p)^2 . p and so on.

Now you don't have to do this next step but I recommend it as practises the sum to infinity of a geometric progression GP.

The sum of all these probabilities is

p is a common factor here, and what is left is a GP with first term 1 and common ratio (1-p)

Use the sum to infinity to simplify this. You should get 1 because eventually all probabilities must occur.

But, back to the problem. The expected value for the number of goes is the sum of (n times P(n)) for all n.

Now that would leave me puzzled except that many years ago I read of a way to sum this. I would never have thought of it as it involves calculus.

First take out the common factor p and call the remaining sum F(p).

Each term of F(p) is individually integrable so work out the integral of F.

The result is a GP so you can work out the sum to infinity. It comes out to a relatively simple expression. Now for the mathematical magic. You differentiate this with respect to p.

Now put back the factor p and you have the required expected value. Evaluate with the value of p you had at the start and you've got your answer.

For me it wasn't a whole number; they often aren't. It just means if you repeated the test many, many times this number is the average number of goes until success.

As a statistician it's then up to you to interpret what it means. I'd say the expected number of goes is between N and N+1.

Hope that helps. Post again to let me know you've got it or if you need more help.

Bob

Mister_JWO · 2022-10-19 05:57:13

Thank you for the method.

I understand your solution a little bit, but I've not learned yet about integrals and differentials.

Is there a simple trick to handle with this?

Bob · 2022-10-19 07:35:42

hi

My program and Norton Security had an argument and Norton won. I had to find out how to exclude my prog and then write it again. Gggrrrrr!

But the good news is the simulation gave good agreement with my method so I'm sticking with it for now.

I think there is a non calculus way to do the last steps but I'll have to research it. Back tomorrow.

Bob

Mister_JWO · 2022-10-19 07:43:39

Thank you very much!

Bob · 2022-10-19 20:58:42

hi Mister_JWO

I have re-written my program to copy exactly what the game show says and run it over 200000 times. The average number of goes agrees with my algebraic calculation to 2 decimal places, so I'm happy with my method.

I have also found a non calculus way to find the sum of terms. I'll start you off here:

This can be split into two series:

The first series is a GP and so the sum to infinity can be calculated. The second has a common factor (1-p) and if that is 'extracted' what remains is F(p) again. So you can re-arrange the equation so that F(p) is given in terms of p. The expected number of attempts is p.F(p).

Bob

zetafunc · 2022-10-20 00:26:27

Just a precautionary note -- while Bob's approach in post #6 is correct, we can only do this because the series converges. In other words, the logic would be:

(1) First prove that the series converges, e.g. by looking at the sequence of partial sums (that's the sum of the first N terms) and showing that that sequence converges as N gets larger.
(2) Now that we know the sum converges, we can then do the rearranging of terms that Bob outlined in post #6 to identify what it converges to.

If the series did not converge then this kind of manipulation is not allowed -- take for example the divergent series 1 - 1 + 1 - 1 + 1 - ... which can be rearranged to 'converge' to lots of different things.

Similar reasoning is also required for differentiating/integrating the infinite series in post #2. We can do this here because p is a probability (so is between 0 and 1, within the series' radius of convergence) and so the series is integrable within that radius of convergence (and its integral also converges).

Math Is Fun Forum

#1 2022-10-19 00:22:24

Extra exercise during class

#2 2022-10-19 04:54:55

Re: Extra exercise during class

#3 2022-10-19 05:57:13

Re: Extra exercise during class

#4 2022-10-19 07:35:42

Re: Extra exercise during class

#5 2022-10-19 07:43:39

Re: Extra exercise during class

#6 2022-10-19 20:58:42

Re: Extra exercise during class

#7 2022-10-20 00:26:27

Re: Extra exercise during class

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