# statistics question

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#1
How do I solve part c? I thought of it along the lines of inverse binomial distribution where-
X~B(120, 5/120)
P(X=x)=0.95
How do I find x?

Any help would be appreciated
Last edited by mmike2003; 3 weeks ago
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3 weeks ago
#2
so firstly the sign should not be equal but smaller or equal, and then you use the cumulative thing in your calc and you should find it that way by looking for the first one which is above 0.95, so in this case 9 i think
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#3
(Original post by dmears55)
so firstly the sign should not be equal but smaller or equal, and then you use the cumulative thing in your calc and you should find it that way by looking for the first one which is above 0.95, so in this case 9 i think
Thank you this was really helpful
1
3 weeks ago
#4
(Original post by dmears55)
so firstly the sign should not be equal but smaller or equal, and then you use the cumulative thing in your calc and you should find it that way by looking for the first one which is above 0.95, so in this case 9 i think
Agree about the sign, but not with the rest of it.

mmike2003

You have an open ended situation here. It is NOT a binomial distribution.
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#5
(Original post by ghostwalker)
Agree about the sign, but not with the rest of it.

mmike2003

You have an open ended situation here. It is NOT a binomial distribution.
Oh wait, I thought I use binomial distribution. Any hint? Or could you show some working for help please? Thanks
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3 weeks ago
#6
(Original post by mmike2003)
Oh wait, I thought I use binomial distribution. Any hint? Or could you show some working for help please? Thanks
It's easiest to look at the complementary event. So, we're looking for the probability of him not winning in n weeks, say, being less than 0.05

If we let p be the probability that he wins in a given week, then:

What's the probability that he doesn't win in the first week?
Ditto in the first two weeks?
And the first three?

See the pattern. Then consider n weeks and form an inequality.
Last edited by ghostwalker; 3 weeks ago
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#7
(Original post by ghostwalker)
It's easiest to look at the complementary event. So, we're looking for the probability of him not winning in n weeks, say, being less than 0.05

If we let p be the probability that he wins in a given week, then:

What's the probability that he doesn't win in the first week?
Ditto in the first two weeks?
And the first three?

See the pattern. Then consider n weeks and form an inequality.
Thanks for the hint.

The probability that he wins on any given week is 5/120
So, the probability that he doesn't win on any given week is 1- [5/120] = 23/24

how do I find week two and three? Once I get these I think I can find the pattern to solve the inequality
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3 weeks ago
#8
(Original post by mmike2003)
Thanks for the hint.

The probability that he wins on any given week is 5/120
So, the probability that he doesn't win on any given week is 1- [5/120] = 23/24

how do I find week two and three? Once I get these I think I can find the pattern to solve the inequality
Don't know the previous answers, so I assume your figures are correct.

Are weeks 2 and 3 any different to week 1, in terms of the probability of not winning?

Note: We're not just interested in just week2, but rather the first two weeks together, etc.
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#9
(Original post by ghostwalker)
Don't know the previous answers, so I assume your figures are correct.

Are weeks 2 and 3 any different to week 1, in terms of the probability of not winning?

Note: We're not just interested in just week2, but rather the first two weeks together, etc.
Yes, I would assume it is given by (23/24)^n , where n is the number of weeks???

so week two would be (23/24)^2=0.918??
and week three would be (23/24)^3=0.8801?

Am I thinking along the right lines?
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3 weeks ago
#10
(Original post by mmike2003)
Yes, I would assume it is given by (23/24)^n , where n is the number of weeks???

so week two would be (23/24)^2=0.918??
and week three would be (23/24)^3=0.8801?

Am I thinking along the right lines?
The probabilty of failing in week2 is still 23/24, so the probablity of not winning in the first two weeks is (23/24)^2

Etc.

And hence the probablity of failing in all the first n weeks (i.e. not winning) is (23/24)^n, as you have.

So, it just remains to form the inequality and solve.
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#11
(Original post by ghostwalker)
The probabilty of failing in week2 is still 23/24, so the probablity of not winning in the first two weeks is (23/24)^2

Etc.

And hence the probablity of failing in all the first n weeks (i.e. not winning) is (23/24)^n, as you have.

So, it just remains to form the inequality and solve.
I need to find the number of weeks to achieve a probability of winning at 95%, hence a probability of not winning at 5%

(23/24)^n<0.05
so, n=log base 23/24 (0.05)= 70.3890 . This rounds to 71 weeks I guess

Is this correct?
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3 weeks ago
#12
(Original post by mmike2003)
I need to find the number of weeks to achieve a probability of winning at 95%, hence a probability of not winning at 5%

(23/24)^n<0.05
so, n=log base 23/24 (0.05)= 70.3890 . This rounds to 71 weeks I guess

Is this correct?
Looks good.

And you should be aware that the distribution you're looking at is the Geometric.
1
#13
(Original post by ghostwalker)
Looks good.

And you should be aware that the distribution you're looking at is the Geometric.
Thank you this was very helpful
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