Do we compute the probability as a binomial distribution?

burgerandcheese

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Question: In a survey carried out in a certain school, it is found that 3 out of 7 students passed the Mathematics test. If 8 students from that school are chosen at random, calculate the probability that

i. Exactly 5 students passed the test
ii. At least 3 students passed the test

--------------------------------------------

I just read back on binomial distribution and it said
"Bernoulli trials have the following three properties:
1. Each trial has only two possible outcomes, success and failure.
2. On every trial, each outcome has a fixed probability of occurring. If a success has a probability of p, then a failure has a probability of 1 - p.
3. The trials are independent of each other. The outcome of one trial has no influence over the outcome of another trial."

So if we do treat this question as a binomial probability, then the probability of each trial must be the same and therefore it's possible to pick the same student multiple times. In other words, the students are chosen with replacement. Does "8 students from that school are chosen at random" imply so?

This is what I did for part (i) using the the Binomial Distribution formula:
P(X = r) = C(n, r) * (p^r) * [q^(n - r)], where r = 0, 1, 2, 3, ..., n and p + q = 1

number of trials, n = 8
probability of success in each trial, p = 3/7
probability of failure in each trial, q = 1 - p = 4/7
number of successful trials, r = 5

P(X = 5) = C(8, 5) * (3/7)⁵ * (4/7)³
..= 0.1511 (4 significant figures)
 
Question: In a survey carried out in a certain school, it is found that 3 out of 7 students passed the Mathematics test. If 8 students from that school are chosen at random, calculate the probability that

i. Exactly 5 students passed the test
ii. At least 3 students passed the test

--------------------------------------------

I just read back on binomial distribution and it said
"Bernoulli trials have the following three properties:
1. Each trial has only two possible outcomes, success and failure.
2. On every trial, each outcome has a fixed probability of occurring. If a success has a probability of p, then a failure has a probability of 1 - p.
3. The trials are independent of each other. The outcome of one trial has no influence over the outcome of another trial."

So if we do treat this question as a binomial probability, then the probability of each trial must be the same and therefore it's possible to pick the same student multiple times. In other words, the students are chosen with replacement. Does "8 students from that school are chosen at random" imply so?

This is what I did for part (i) using the the Binomial Distribution formula:
P(X = r) = C(n, r) * (p^r) * [q^(n - r)], where r = 0, 1, 2, 3, ..., n and p + q = 1

number of trials, n = 8
probability of success in each trial, p = 3/7
probability of failure in each trial, q = 1 - p = 4/7
number of successful trials, r = 5

P(X = 5) = C(8, 5) * (3/7)⁵ * (4/7)³
..= 0.1511 (4 significant figures)
And then what is your question?
 
Question: In a survey carried out in a certain school, it is found that 3 out of 7 students passed the Mathematics test. If 8 students from that school are chosen at random, calculate the probability that

i. Exactly 5 students passed the test
ii. At least 3 students passed the test

--------------------------------------------

I just read back on binomial distribution and it said
"Bernoulli trials have the following three properties:
1. Each trial has only two possible outcomes, success and failure.
2. On every trial, each outcome has a fixed probability of occurring. If a success has a probability of p, then a failure has a probability of 1 - p.
3. The trials are independent of each other. The outcome of one trial has no influence over the outcome of another trial."

So if we do treat this question as a binomial probability, then the probability of each trial must be the same and therefore it's possible to pick the same student multiple times. In other words, the students are chosen with replacement. Does "8 students from that school are chosen at random" imply so?

This is what I did for part (i) using the the Binomial Distribution formula:
P(X = r) = C(n, r) * (p^r) * [q^(n - r)], where r = 0, 1, 2, 3, ..., n and p + q = 1

number of trials, n = 8
probability of success in each trial, p = 3/7
probability of failure in each trial, q = 1 - p = 4/7
number of successful trials, r = 5

P(X = 5) = C(8, 5) * (3/7)⁵ * (4/7)³
..= 0.1511 (4 significant figures)
P(X = k) = C(8, k) * (3/7)k * (4/7)8-k
 
Question: In a survey carried out in a certain school, it is found that 3 out of 7 students passed the Mathematics test. If 8 students from that school are chosen at random, calculate the probability that

i. Exactly 5 students passed the test
ii. At least 3 students passed the test

--------------------------------------------

I just read back on binomial distribution and it said
"Bernoulli trials have the following three properties:
1. Each trial has only two possible outcomes, success and failure.
2. On every trial, each outcome has a fixed probability of occurring. If a success has a probability of p, then a failure has a probability of 1 - p.
3. The trials are independent of each other. The outcome of one trial has no influence over the outcome of another trial."

So if we do treat this question as a binomial probability, then the probability of each trial must be the same and therefore it's possible to pick the same student multiple times. In other words, the students are chosen with replacement. Does "8 students from that school are chosen at random" imply so?
No. It is not "student being chosen" or not that is the random variable here. It is "student passes test" or not.
If the size of the school were small compared to the size of the sample, then this not be a binomial distribution. However, here, we are not told the size of the school so we must assume it is large.

This is what I did for part (i) using the the Binomial Distribution formula:
P(X = r) = C(n, r) * (p^r) * [q^(n - r)], where r = 0, 1, 2, 3, ..., n and p + q = 1

number of trials, n = 8
probability of success in each trial, p = 3/7
probability of failure in each trial, q = 1 - p = 4/7
number of successful trials, r = 5

P(X = 5) = C(8, 5) * (3/7)⁵ * (4/7)³
..= 0.1511 (4 significant figures)
 
Alright. Thanks guys! By the way, I'm still new here, so please tell me if it's unnecessary to reply just to say thanks, it makes my thread go on top..
 
Alright. Thanks guys! By the way, I'm still new here, so please tell me if it's unnecessary to reply just to say thanks, it makes my thread go on top..
It is courteous to do so but not necessary. What is objected to is bumping to the top to try to get a quicker answer.
 
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