1. ## Probability

omg i feel stupid for asking this question cause I cant get it, im probs tired or something, but here it is anyways:

One layer of tinting material on a window cuts out 1/5 of the sun's UV rays.
1)what fraction would be cut out by using two layers?
2) How many layers would be required to cut out at least 9/10 of the sun's UV rays?

1)9/25
2)11

SHOWS me workings!! fanks

2. ## Re: Probability

1) after 1 layer , we have 1-0.2 =0.8 passing thru
so 0.8 get to travel to the 2nd layer, where another 0.2 will be cut out

2) so it works in the same way as compound interest , only difference is that it "depreciates"
and a similar example would be
if the interest rate is 5% p a, how many years would a principal of P double if the interest if compounded annually ?

so P*(1.05) is the money after 1 year, P*(1.05)*(1.05) is the money after another year
after n years , we have P*1.05^n which should equal 2*P
so 1.05^n=2
take log of both sides
ln (1.05^n)= ln2
n *ln 1.05= ln2
n = ln2 / ln 1.05 = 14.2066...=15 for it to double as it only calculates interest annually

The question you're asking is very similar, but it has inequality in it
so be carefully with the inequality signs

3. ## Re: Probability

also if u haven't learned log , sub number in and see when it satisfies the inequality

which sucks when the number is large

4. ## Re: Probability

Originally Posted by frmldhyd
omg i feel stupid for asking this question cause I cant get it, im probs tired or something, but here it is anyways:

One layer of tinting material on a window cuts out 1/5 of the sun's UV rays.
1)what fraction would be cut out by using two layers?
2) How many layers would be required to cut out at least 9/10 of the sun's UV rays?

1)9/25
2)11

SHOWS me workings!! fanks
This is a series type question, not probability.
Let x be the amount of the sun's rays that hits the material
First layer takes out x/5 so after the first layer there is only 4x/5 left that is not cut
When it hits the second layer, only 4x/5 of rays pass through and a fifth of that is removed
i.e. (4x/5)(1/5) is removed
Total amount cut = x/5 + 4x/25 = 9x/25 so fraction of rays cut is 9/25

Note that after each layer we multiply the amount of rays that entered by 4/5 since 1/5 is removed.
In terms of proportions:
After 1st layer the fraction of rays left is (4/5)
After 2nd layer the fraction of rays left is (4/5)2
After 3rd layer the fraction of rays left is (4/5)3
...
After nth layer the fraction of rays left is (4/5)n

We want the fraction of rays left after the nth layer to be at least 1/10 (as 9/10 is cut out) so
(4/5)n > 9/10
Using logarithms or otherwise this should yield n = 11 as the lowest integer which satisfies this inequality

Thanks!!

6. ## Re: Probability

Originally Posted by Trebla
This is a series type question, not probability.
LOL, I felt that something was not right

7. ## Re: Probability

Originally Posted by Trebla
This is a series type question, not probability.
Let x be the amount of the sun's rays that hits the material
First layer takes out x/5 so after the first layer there is only 4x/5 left that is not cut
When it hits the second layer, only 4x/5 of rays pass through and a fifth of that is removed
i.e. (4x/5)(1/5) is removed
Total amount cut = x/5 + 4x/25 = 9x/25 so fraction of rays cut is 9/25

Note that after each layer we multiply the amount of rays that entered by 4/5 since 1/5 is removed.
In terms of proportions:
After 1st layer the fraction of rays left is (4/5)
After 2nd layer the fraction of rays left is (4/5)2
After 3rd layer the fraction of rays left is (4/5)3
...
After nth layer the fraction of rays left is (4/5)n

We want the fraction of rays left after the nth layer to be at least 1/10 (as 9/10 is cut out) so
(4/5)n > 9/10
Using logarithms or otherwise this should yield n = 11 as the lowest integer which satisfies this inequality
i believe you have a made a mistake but please correct me if i am wrong
where you wrote (4/5)^n>9/10, it is meant to be (4/5)^n<1/10 as the number of layers you want to cut out is 9/10 and leave 1/10 light remaining
thanks

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