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Maths question HELP!!!
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Bazyli
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Hi, I’m really struggling getting my head around this question (last question on S1 OCR June 2009). Please can someone help point me in the right direction? Thanks
Repeated independent trials of a certain experiment are carried out. on each trial the probability of success is 0.12.
a) Find the smallest value of n such that the probability of at least one least one success in n trials is more than 0.95
b) Find the probability that the 3rd success occurs on the 7th trial
Repeated independent trials of a certain experiment are carried out. on each trial the probability of success is 0.12.
a) Find the smallest value of n such that the probability of at least one least one success in n trials is more than 0.95
b) Find the probability that the 3rd success occurs on the 7th trial
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mqb2766
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The opposite of a) is that all trials fail.
(Original post by Bazyli)
Hi, I’m really struggling getting my head around this question (last question on S1 OCR June 2009). Please can someone help point me in the right direction? Thanks
Repeated independent trials of a certain experiment are carried out. on each trial the probability of success is 0.12.
a) Find the smallest value of n such that the probability of at least one least one success in n trials is more than 0.95
b) Find the probability that the 3rd success occurs on the 7th trial
Hi, I’m really struggling getting my head around this question (last question on S1 OCR June 2009). Please can someone help point me in the right direction? Thanks
Repeated independent trials of a certain experiment are carried out. on each trial the probability of success is 0.12.
a) Find the smallest value of n such that the probability of at least one least one success in n trials is more than 0.95
b) Find the probability that the 3rd success occurs on the 7th trial
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liam.ro
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a) Firstly, identify the most suitable distribution to model the number of successes in n trials (random variable X). Once you have that, you need to go through trial and error to find the smallest value of n for which P(X≥1) in your distribution.
b) The premise of this subquestion hasn't changed, so you will be using the same distribution (although there is an alternative you may like to use). If you're continuing with this distribution, here's a hint: having a 3rd success on the 7th trial is equivalent to having 2 successes in the first 6 trials, followed by a success on the next trial.
As I've intentionally been a bit obscure with my guidance, let me know if you're struggling!
b) The premise of this subquestion hasn't changed, so you will be using the same distribution (although there is an alternative you may like to use). If you're continuing with this distribution, here's a hint: having a 3rd success on the 7th trial is equivalent to having 2 successes in the first 6 trials, followed by a success on the next trial.
As I've intentionally been a bit obscure with my guidance, let me know if you're struggling!
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Bazyli
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#4
(Original post by liam.ro)
a) Firstly, identify the most suitable distribution to model the number of successes in n trials (random variable X). Once you have that, you need to go through trial and error to find the smallest value of n for which P(X≥1) in your distribution.
b) The premise of this subquestion hasn't changed, so you will be using the same distribution (although there is an alternative you may like to use). If you're continuing with this distribution, here's a hint: having a 3rd success on the 7th trial is equivalent to having 2 successes in the first 6 trials, followed by a success on the next trial.
As I've intentionally been a bit obscure with my guidance, let me know if you're struggling!
a) Firstly, identify the most suitable distribution to model the number of successes in n trials (random variable X). Once you have that, you need to go through trial and error to find the smallest value of n for which P(X≥1) in your distribution.
b) The premise of this subquestion hasn't changed, so you will be using the same distribution (although there is an alternative you may like to use). If you're continuing with this distribution, here's a hint: having a 3rd success on the 7th trial is equivalent to having 2 successes in the first 6 trials, followed by a success on the next trial.
As I've intentionally been a bit obscure with my guidance, let me know if you're struggling!
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liam.ro
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Unfortunately, this situation cannot be modelled by a geometric distribution. It can only be modelled by either a binomial or negative binomial distribution. (You probably chose the geometric distribution because it's actually a special case of the negative binomial distribution where r=1; so intuitively, they seem similar.)
Binomial Distribution:
As I previously stated, having a 3rd success on the 7th trial is equivalent to having 2 successes in 6 trials, followed by a success in the next trial. The probability of the first part can be modelled binomially, and the second part is just your p value, giving your total probability as their product.
Negative Binomial Distribution:
In case you're not familiar with it, the negative binomial distribution is modelled by X~NB(r,p) where X is the number of trials required to achieve r successes, each with probability p. If you use this model, the calculation would only require plugging values into the negative binomial probability formula, which you can easily find online.
Both methods should get you to the same, hopefully correct, answer. I always found the binomial method to be much more intuitive, but it clicks differently for everyone. Let me know how you get on!
Binomial Distribution:
As I previously stated, having a 3rd success on the 7th trial is equivalent to having 2 successes in 6 trials, followed by a success in the next trial. The probability of the first part can be modelled binomially, and the second part is just your p value, giving your total probability as their product.
Negative Binomial Distribution:
In case you're not familiar with it, the negative binomial distribution is modelled by X~NB(r,p) where X is the number of trials required to achieve r successes, each with probability p. If you use this model, the calculation would only require plugging values into the negative binomial probability formula, which you can easily find online.
Both methods should get you to the same, hopefully correct, answer. I always found the binomial method to be much more intuitive, but it clicks differently for everyone. Let me know how you get on!
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Bazyli
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#6
Thanks for all your help. I’ve found the answer by using a binomial distribution: 6C2xp^2xq^4 and then multiplying by p to get the 3rd success🙂
(Original post by liam.ro)
Unfortunately, this situation cannot be modelled by a geometric distribution. It can only be modelled by either a binomial or negative binomial distribution. (You probably chose the geometric distribution because it's actually a special case of the negative binomial distribution where r=1; so intuitively, they seem similar.)
Binomial Distribution:
As I previously stated, having a 3rd success on the 7th trial is equivalent to having 2 successes in 6 trials, followed by a success in the next trial. The probability of the first part can be modelled binomially, and the second part is just your p value, giving your total probability as their product.
Negative Binomial Distribution:
In case you're not familiar with it, the negative binomial distribution is modelled by X~NB(r,p) where X is the number of trials required to achieve r successes, each with probability p. If you use this model, the calculation would only require plugging values into the negative binomial probability formula, which you can easily find online.
Both methods should get you to the same, hopefully correct, answer. I always found the binomial method to be much more intuitive, but it clicks differently for everyone. Let me know how you get on!
Unfortunately, this situation cannot be modelled by a geometric distribution. It can only be modelled by either a binomial or negative binomial distribution. (You probably chose the geometric distribution because it's actually a special case of the negative binomial distribution where r=1; so intuitively, they seem similar.)
Binomial Distribution:
As I previously stated, having a 3rd success on the 7th trial is equivalent to having 2 successes in 6 trials, followed by a success in the next trial. The probability of the first part can be modelled binomially, and the second part is just your p value, giving your total probability as their product.
Negative Binomial Distribution:
In case you're not familiar with it, the negative binomial distribution is modelled by X~NB(r,p) where X is the number of trials required to achieve r successes, each with probability p. If you use this model, the calculation would only require plugging values into the negative binomial probability formula, which you can easily find online.
Both methods should get you to the same, hopefully correct, answer. I always found the binomial method to be much more intuitive, but it clicks differently for everyone. Let me know how you get on!
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