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How would you find p in this equation?



the solutions book just says, ''solve p by iteration to give 0.027''

does 'iteration' mean you have to stick in numbers by trial and error until you get the right answer? :s-smilie:

theres got to be a simpler way surely
How many marks is the question?

I would assume they mean to substitute in numbers around 0.27 to 3dp, e.g 0.265 and 0.275 so show that to 2dp it equals 0.27.
Reply 2
Avatar for Jam'
Jam'
15
(1+19p)/(1-p)=-1/19

So (1+19p)(1-p)^19 = -1/19 (1-p)^20 = 0.9
(edited 9 years ago)
Reply 3
Avatar for Red Richie
Red Richie
OP
15
Original post by Jam'
(1+19p)/(1-p)=-19

So (1+19p)(1-p)^19 = -19 (1-p)^20


its (1+19p)/(1-p)= 0.9
Reply 4
Avatar for Red Richie
Red Richie
OP
15
Original post by acciolucy
How many marks is the question?

I would assume they mean to substitute in numbers around 0.27 to 3dp, e.g 0.265 and 0.275 so show that to 2dp it equals 0.27.


its just a textbook question, so no stated marks

so are they basically asking me to do trial and error to find p?
Reply 5
Avatar for Jam'
Jam'
15
Original post by Red Richie
its (1+19p)/(1-p)= 0.9


Sorry my mistake but it's not 0.9 either. It's -1/19

Do you know how to do long division?
Reply 6
Avatar for Red Richie
Red Richie
OP
15
Original post by Jam'
Sorry my mistake but it's not 0.9 either. It's -1/19

Do you know how to do long division?


Let me post the full question and the solution, its actually a probability question:

A machine makes components, and the probability that a component is defective is p. If the components are packed in cartons of 20, what value of p will ensure that 90% of cartons contain at most one defective component?

solution:



0.9 is the given probability in the question

yeah I know long division, haven't done it for ages though, how would I use it to find p?
(edited 9 years ago)
Original post by Red Richie
...


Iteration, means use numerical methods. E.g. you could rearrange your equation to be.

pn+1=1−0.91+19pn19\displaystyle p_{n+1}=1-\sqrt[19]{\frac{0.9}{1+19p_n}}

Choose a reasonable starting point, say p0=0.5p_0=0.5 and keep plugging in the new value to the RHS, until it's converged sufficiently.

PS: Not checked whether it actually converges, but seems a reasonable equation to use.
(edited 9 years ago)
Original post by ghostwalker
Iteration, means use numerical methods. E.g. you could rearrange your equation to be.

pn+1=1−0.91+19pn19\displaystyle p_{n+1}=1-\sqrt[19]{\frac{0.9}{1+19p_n}}

Choose a reasonable starting point, say p0=0.5p_0=0.5 and keep plugging in the new value to the RHS, until it's converged sufficiently.

PS: Not checked whether it actually converges, but seems a reasonable equation to use.


It does converge but desperately slowly.

This is a bit better.

Unparseable latex formula:

p_{n+1}=1-\sqrt[19] {\dfrac{0.9 - (1 - p_n)^{19}}{19 p_n}}}



Challenge: Find a better iterative formula by rearrangement (no differentiation allowed).
(edited 9 years ago)
Original post by Mr M


Challenge: Find a better iterative formula by rearrangement (no differentiation allowed).


Fail :cry:

My other attempts diverged!
Original post by ghostwalker
Fail :cry:

My other attempts diverged!


You can do it. Show resilience!
Original post by Mr M
You can do it. Show resilience!


OK

pn+1=1−0.9−(1−pn)2020pn19\displaystyle p_{n+1}=1-\sqrt[19]{\frac{0.9-(1-p_n)^{20}}{20p_n}}

Not that dissimilar to yours
Original post by ghostwalker
OK

pn+1=1−0.9−(1−pn)2020pn19\displaystyle p_{n+1}=1-\sqrt[19]{\frac{0.9-(1-p_n)^{20}}{20p_n}}

Not that dissimilar to yours


It's a goodie though.

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