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# Confusion in Probability

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Manager
Joined: 08 Apr 2019
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Location: India
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09 Jun 2019, 22:42
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?

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10 Jun 2019, 00:30
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/

or if you prefer your answers in video format, then the link below might be useful too-

please let me know if you have any question, would be happy to help
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10 Jun 2019, 13:59
2
rishabhjain13 wrote:
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?

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.
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My latest GMAT blog posts | Suggestions for blog articles are always welcome!
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11 Jun 2019, 02:53
rishabhjain13 wrote:
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?

When you have to form a team or a group, you don't have to worry about the order in which the team members are selected. A team GGBBBB is the same as team GBGBBB which is the same as BBBBGG etc. This is just one instance of a team. You just need to select 2 girls in 6C2 ways and select 4 boys in 6C4 ways (to decide which two girls and which 2 boys will be part of the team)

In question 2, think about it for a moment - what is the probability that when you flip a coin 8 times, you will get heads on first 3 flips and tails on last 5 flips?
(1/2)*(1/2)*(1/2) * (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = (0.5)^3 * (0.5)^5 = .0039
This is the probability of getting HHHTTTTT.

But we don't want the probability of HHHTTTTT. We want the probability of all instances such that there are exactly 3 heads. So probability of HTTHHTTT needs to be added to it. Probability of TTHTHTHT needs to be added to it and so on... Probability of each of these cases will also be .0039

So you multiply .0039 by 8C3 i.e. the number of ways in which you can have 3 Hs.

Sometimes, working with probability can be confusing as to whether we have to arrange or not. Hence, I always suggest to work with cases only.
i.e. Total cases = 2*2*2*2*2*2*2*2 = 2^8

Favourable cases = 8C3 (select 3 flips and put heads there. Rest all have tails)

Req Probability = 8C3 / 2^8

Check out this post too: https://www.veritasprep.com/blog/2013/0 ... er-matter/
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Updated on: 11 Jun 2019, 19:13
ccooley wrote:
rishabhjain13 wrote:
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?

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.

Thank you very much, ccooley. The explanation is definitely helpful.

Originally posted by RJ7X0DefiningMyX on 11 Jun 2019, 19:10.
Last edited by RJ7X0DefiningMyX on 11 Jun 2019, 19:13, edited 1 time in total.
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11 Jun 2019, 19:12
rishabhjain13 wrote:
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?

When you have to form a team or a group, you don't have to worry about the order in which the team members are selected. A team GGBBBB is the same as team GBGBBB which is the same as BBBBGG etc. This is just one instance of a team. You just need to select 2 girls in 6C2 ways and select 4 boys in 6C4 ways (to decide which two girls and which 2 boys will be part of the team)

In question 2, think about it for a moment - what is the probability that when you flip a coin 8 times, you will get heads on first 3 flips and tails on last 5 flips?
(1/2)*(1/2)*(1/2) * (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = (0.5)^3 * (0.5)^5 = .0039
This is the probability of getting HHHTTTTT.

But we don't want the probability of HHHTTTTT. We want the probability of all instances such that there are exactly 3 heads. So probability of HTTHHTTT needs to be added to it. Probability of TTHTHTHT needs to be added to it and so on... Probability of each of these cases will also be .0039

So you multiply .0039 by 8C3 i.e. the number of ways in which you can have 3 Hs.

Sometimes, working with probability can be confusing as to whether we have to arrange or not. Hence, I always suggest to work with cases only.
i.e. Total cases = 2*2*2*2*2*2*2*2 = 2^8

Favourable cases = 8C3 (select 3 flips and put heads there. Rest all have tails)

Req Probability = 8C3 / 2^8

Check out this post too: https://www.veritasprep.com/blog/2013/0 ... er-matter/

Thank you very much, VeritasKarishma. Yes, at times, I do prefer to work with cases.
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27 Jun 2019, 01:21
A man is standing in a square cell on chess board. Now he can move/step straight either to the right, left, front and back at each movement/step. The probability of every movement is same. So what is the probability that he'll be back to his initial position after 9th step?

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27 Jun 2019, 01:57
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HossainAlam01 wrote:
A man is standing in a square cell on chess board. Now he can move/step straight either to the right, left, front and back at each movement/step. The probability of every movement is same. So what is the probability that he'll be back to his initial position after 9th step?

Posted from my mobile device

The probability of reaching his initial position after the 9th step would be ZERO, since it would require the man an 'even' number of moves to reach back to his initial position.
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27 Jun 2019, 02:17
The probability of reaching his initial position after the 9th step would be ZERO, since it would require the man an 'even' number of moves to reach back to his initial position.

Posted from my mobile device
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27 Jun 2019, 02:22
HossainAlam01 wrote:
The probability of reaching his initial position after the 9th step would be ZERO, since it would require the man an 'even' number of moves to reach back to his initial position.

Posted from my mobile device

I will do that. However, please share the complete list of all options with me as well. That will help me in making you understand the solution better.
Re: Confusion in Probability   [#permalink] 27 Jun 2019, 02:22
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