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09 Jun 2019, 22:42
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I have confusion in the following two types of questions:
Type 1: A team of 6 is to be formed from a group of 6 girls and 6 boys. Find the probability of making a team that consists of at least two girls.
-We use the method of calculating the probability of 2 Girls and 4 Boys + probability of 3 Girls and 3 Boys + probability of 4 Girls and 2 Boys + probability of 5 Girls and 1 Boy + probability of 6 Girls and 0 Boys (or do it by the Non-Event method, i.e. probability of 0 Girls and 6 Boys + probability of 1 Girl and 5 Boys)
WHEREAS, in question Type 2: Probability of getting 3 heads while tossing a coin 8 times?
-We use something like 8C3 * (0.5)^3 * (0.5)^5
Using 8C3 to determine 3 tosses where Heads will appear
Now, my question is, in Type 1, why don't we determine the slots that the girls would take? I know, that the arrangement doesn't matter since we only have to 'select' girls and not arrange them - BUT THEN, in Type 2, why are we then concerned about that? Why can't we just keep it like (0.5)^3 * (0.5)^5?
I know, this is a little confusing, and perhaps, even my question isn't very much clear, but broadly, why do we need 8C3 in Type 2 is what I want to know?
Bunuel VeritasKarishma EgmatQuantExpert
Type 1: A team of 6 is to be formed from a group of 6 girls and 6 boys. Find the probability of making a team that consists of at least two girls.
-We use the method of calculating the probability of 2 Girls and 4 Boys + probability of 3 Girls and 3 Boys + probability of 4 Girls and 2 Boys + probability of 5 Girls and 1 Boy + probability of 6 Girls and 0 Boys (or do it by the Non-Event method, i.e. probability of 0 Girls and 6 Boys + probability of 1 Girl and 5 Boys)
WHEREAS, in question Type 2: Probability of getting 3 heads while tossing a coin 8 times?
-We use something like 8C3 * (0.5)^3 * (0.5)^5
Using 8C3 to determine 3 tosses where Heads will appear
Now, my question is, in Type 1, why don't we determine the slots that the girls would take? I know, that the arrangement doesn't matter since we only have to 'select' girls and not arrange them - BUT THEN, in Type 2, why are we then concerned about that? Why can't we just keep it like (0.5)^3 * (0.5)^5?
I know, this is a little confusing, and perhaps, even my question isn't very much clear, but broadly, why do we need 8C3 in Type 2 is what I want to know?
Bunuel VeritasKarishma EgmatQuantExpert
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sakshamchhabra
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10 Jun 2019, 00:30
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Greetings rishabhjain13,
I am no expert but here are my 2 cents,
Rishabh in the solution to the second question, you have used the binomial theorem for probability.
From my point of view, the reason for your confusion is that both of your questions are very different. Furthermore, the answer in your second question is a clear-cut formula application.
The binomial theorem is an easier way to determine the probability of doing something x number of times out of total n number of ways.
The formula for binomial probability is-\(nCx (p)^x (q)^{n-x}\)
here, n= total outcomes; x = favourable; p = probabilty of desirable outcome; q= probability of non-desirable outcome - we can calculate it as q = 1-p
There would've been a total of 256 outcomes ( 2^8), and we would have to find all the cases where heads occur 3 times so to avoid such uphill calculation we can use the formula.
https://www.veritasprep.com//blog/2012/02/quarter-wit-quarter-wisdom-braving-the-binomial-probability/
this article might be useful to better understand the scenario.
or if you prefer your answers in video format, then the link below might be useful too-
https://www.khanacademy.org/math/precalculus/prob-comb/prob-combinatorics-precalc/v/generalizing-with-binomial-coefficients-bit-advanced
please let me know if you have any question, would be happy to help
I am no expert but here are my 2 cents,
Rishabh in the solution to the second question, you have used the binomial theorem for probability.
From my point of view, the reason for your confusion is that both of your questions are very different. Furthermore, the answer in your second question is a clear-cut formula application.
The binomial theorem is an easier way to determine the probability of doing something x number of times out of total n number of ways.
The formula for binomial probability is-\(nCx (p)^x (q)^{n-x}\)
here, n= total outcomes; x = favourable; p = probabilty of desirable outcome; q= probability of non-desirable outcome - we can calculate it as q = 1-p
There would've been a total of 256 outcomes ( 2^8), and we would have to find all the cases where heads occur 3 times so to avoid such uphill calculation we can use the formula.
https://www.veritasprep.com//blog/2012/02/quarter-wit-quarter-wisdom-braving-the-binomial-probability/
this article might be useful to better understand the scenario.
or if you prefer your answers in video format, then the link below might be useful too-
https://www.khanacademy.org/math/precalculus/prob-comb/prob-combinatorics-precalc/v/generalizing-with-binomial-coefficients-bit-advanced
please let me know if you have any question, would be happy to help
10 Jun 2019, 13:59
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rishabhjain13
The difference is whether 'order matters.'
The easiest way to tell whether you're doing an 'order matters' problem, or an 'order doesn't matter' problem, is by thinking about which scenarios you want to count as being the same.
In the coin flip scenario, flipping H H H T T T T T is a different situation from flipping T T H T H H T T, etc. Even though there are the same number of heads in both of those scenarios, they're different sets of coin flips. You wouldn't say that H H H T T T T T is exactly the same series of coin flips as T T H T H H T T: you'd say that they're different, but they have the same number of heads.
But 'Sue, Harriet, Maya, Gretchen, Claire, and Jo' is the SAME team as 'Harriet, Sue, Gretchen, Maya, Jo, and Claire.' So you don't want to count those different orders as being different. You wouldn't say that those were two different groups of people; at most, you'd say that they're the same group but written differently.
So, in the coin flip scenario, we care about counting outcomes with the heads in different 'slots' as different outcomes. In the scenario with the people, we don't care about that.
