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junior college 2 | H2 Maths
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Pls help me with (iv) using binomial distribution. Thanks
Date Posted: 4 years ago
Views: 503
This one is a binomial distribution where “success rate” = 75% or 0.75.
Then the “failure” rate is 0.25.
The idea is very closely linked to a binomial expansion of (a + b)^n where the a and b represents some sort of success and failure; each term in the expansion represents the probability of getting m successes and k failures out of (m + k) trials.
Will look at this tonight if no one answers them yet.
Then the “failure” rate is 0.25.
The idea is very closely linked to a binomial expansion of (a + b)^n where the a and b represents some sort of success and failure; each term in the expansion represents the probability of getting m successes and k failures out of (m + k) trials.
Will look at this tonight if no one answers them yet.
Okay, thanks!
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First three is something like this.
Date Posted:
4 years ago
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The fourth question has three cases to consider.
Date Posted:
4 years ago
Is there an alternative shorter method to consider?
In such cases, we have no choice but to list out the different cases one by one, because each term in the binomial expansion represents one event (a fixed number of successes in this case).
For part ii, if we were to have at least two successes but fewer than five successes, we would have to list out the cases separately; the binomial distribution does not allow for collective results of our wanted value.
For part iv, the only way for us to do this is to fix the fourth event as a success (because the second success HAS to lie in the fourth event; note that had there been two or three success in the first three events, then the success on the fourth event is not the second success already). We then proceed to realise that the first success can lie in any of the first three events, but only exactly one of these three events will have a success in it (resulting in two failures).
I am not sure if your graphing calculator allows for calculation of binomial distribution (my graphing calculator has stopped functioning for about 10 years already), but using a non-graphing approach would require me to list out the cases (since there is no in built function available for non-graphing calculators).
Of course, in the binomial expansion, each term represents the probabilities of each case happening. If you add them, the probability adds up to 1. After all, [p + 1 - p]^n = 1^n = 1.
For part ii, if we were to have at least two successes but fewer than five successes, we would have to list out the cases separately; the binomial distribution does not allow for collective results of our wanted value.
For part iv, the only way for us to do this is to fix the fourth event as a success (because the second success HAS to lie in the fourth event; note that had there been two or three success in the first three events, then the success on the fourth event is not the second success already). We then proceed to realise that the first success can lie in any of the first three events, but only exactly one of these three events will have a success in it (resulting in two failures).
I am not sure if your graphing calculator allows for calculation of binomial distribution (my graphing calculator has stopped functioning for about 10 years already), but using a non-graphing approach would require me to list out the cases (since there is no in built function available for non-graphing calculators).
Of course, in the binomial expansion, each term represents the probabilities of each case happening. If you add them, the probability adds up to 1. After all, [p + 1 - p]^n = 1^n = 1.
Thanks!
When n=3, X~B(3,0.75)
The req prob = P(X=1) x 0.75 x 1 = 0.105
The above is the method from my teacher but i dont understand. Could you help me to explain?
The req prob = P(X=1) x 0.75 x 1 = 0.105
The above is the method from my teacher but i dont understand. Could you help me to explain?
I forgot 99.9% of my stats already actually...
Ok, not really 99;9%, but probably 75%.
What your teacher did is to probably simplify the problem from a "obtaining a second bull's eye in the fourth round" to "obtaining ONE bull's eye in the first three rounds", because after all, they are equivalent (but we must not forget that the fourth one needs to be a success). The story either way represents the same thing.
In such a way, your teacher has redefined the problem into a new problem like this.
"A marksman, ..., scores bull's eyes on 3 out of 4 occasions. He fires THREE rounds at the target...Find the probability that he hits the bull's eye on exactly one of the first three occasions."
After all, if the SECOND bull's eye is on the fourth occasion, then the FIRST must lie somewhere within the first three occasions.
As such, the number of trials for the first three occasions is given by "n" = 3 in this new scenario as your teacher has written. The probability remains 0.75 and remains independent between trials and you can only have two outcomes of success (bull's eye) and failure (off target), which is why a binomial modelling works well here.
X ~ B (3, 0.75) is simply a binomial modelling with 3 trials and a success rate of 75%, as I explained in the previous paragraph.
Probability of obtaining 1 bull's eye is the probability that the number of successes, X, equals 1, also written as P (X = 1).
Of course, we must not forget that the fourth occasion itself is a probabilistic event and we have to take this into account. Hence, we multiply P (X = 1) with the fourth success of 0.75. The last multiplication by 1, I am not sure where it comes from.
As such, probability of hitting a second bull's eye on the fourth occasion
= P (first bull's eye in the first three occasions) * P (bull's eye on the fourth occasion)
= P (X = 1) * 0.75
where P (X = 1) is the probability of getting one success within the first three occasions.
Ok, not really 99;9%, but probably 75%.
What your teacher did is to probably simplify the problem from a "obtaining a second bull's eye in the fourth round" to "obtaining ONE bull's eye in the first three rounds", because after all, they are equivalent (but we must not forget that the fourth one needs to be a success). The story either way represents the same thing.
In such a way, your teacher has redefined the problem into a new problem like this.
"A marksman, ..., scores bull's eyes on 3 out of 4 occasions. He fires THREE rounds at the target...Find the probability that he hits the bull's eye on exactly one of the first three occasions."
After all, if the SECOND bull's eye is on the fourth occasion, then the FIRST must lie somewhere within the first three occasions.
As such, the number of trials for the first three occasions is given by "n" = 3 in this new scenario as your teacher has written. The probability remains 0.75 and remains independent between trials and you can only have two outcomes of success (bull's eye) and failure (off target), which is why a binomial modelling works well here.
X ~ B (3, 0.75) is simply a binomial modelling with 3 trials and a success rate of 75%, as I explained in the previous paragraph.
Probability of obtaining 1 bull's eye is the probability that the number of successes, X, equals 1, also written as P (X = 1).
Of course, we must not forget that the fourth occasion itself is a probabilistic event and we have to take this into account. Hence, we multiply P (X = 1) with the fourth success of 0.75. The last multiplication by 1, I am not sure where it comes from.
As such, probability of hitting a second bull's eye on the fourth occasion
= P (first bull's eye in the first three occasions) * P (bull's eye on the fourth occasion)
= P (X = 1) * 0.75
where P (X = 1) is the probability of getting one success within the first three occasions.
To put it more plainly, your teacher has modelled a bull's eye in the first three occasions as an entirely separate binomial distribution problem (but having the same probability rate of 0.75), which itself is a correct approach.
I practically did what your teacher did, but I listed out the cases fully (there are three cases an I wrote all of them), but in fact it was actually not necessary for me to list out the three cases as it does not actually matter where the first bull's eye landed.
I practically did what your teacher did, but I listed out the cases fully (there are three cases an I wrote all of them), but in fact it was actually not necessary for me to list out the three cases as it does not actually matter where the first bull's eye landed.
Thank you for such a detailed explanation!
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