Problem Solving

Why does OG-11 PS Q-217 calculate prob. with only one possibility? That is, it assumes that the first student will be selected from the junior class with the porb. of 60/1000, and the prob. of selecting 2nd student will then be 1/800. Why not it takes into account the other possibility that first student is selected from senior class with the prob. of 60/800 and then the prob. of 2nd will be 1/1000. And then add the two possibilities to reach to a final prob. After all prob. is all about adding all the possibilities.

Pl. guide.

Nauman

Post the question please. Do not expect people to look up the question and then answer.

My appologies for this. I was kind of confused if i could post the OG material here. So just as a precaution i didnt.

Here's the Question:-

A certain junior class has 1,000 students and a certain senior class has 800 students. Among these srudents, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the porbability that the 2 students selected will be a sibling pair?

A) 3/40,000
B) 1/3,600
C) 9/2,000
D) 1/60
E) 1/15

OA is A.

My question:

Why does OG calculate prob. with only one possibility? That is, it assumes that the first student will be selected from the junior class with the porb. of 60/1000, and the prob. of selecting 2nd student will then be 1/800. Why not it takes into account the other possibility that first student is selected from senior class with the prob. of 60/800 and then the prob. of 2nd will be 1/1000. And then add the two possibilities to reach to a final prob. After all prob. is all about adding all the possibilities.

Pl. guide.

Nauman

is OG wrong on this? can anyone pls explain if the previous post in right?

snuman wrote:My appologies for this. I was kind of confused if i could post the OG material here. So just as a precaution i didnt.

Here's the Question:-

A certain junior class has 1,000 students and a certain senior class has 800 students. Among these srudents, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the porbability that the 2 students selected will be a sibling pair?

A) 3/40,000
B) 1/3,600
C) 9/2,000
D) 1/60
E) 1/15

OA is A.

My question:

Why does OG calculate prob. with only one possibility? That is, it assumes that the first student will be selected from the junior class with the porb. of 60/1000, and the prob. of selecting 2nd student will then be 1/800. Why not it takes into account the other possibility that first student is selected from senior class with the prob. of 60/800 and then the prob. of 2nd will be 1/1000. And then add the two possibilities to reach to a final prob. After all prob. is all about adding all the possibilities.

Pl. guide.

Nauman

I dont think it matters here. If we calculate the probability the way you are calculating - the calculation shall be

[(1/2)*(60/1000)*(1/800) ]+[(1/2)*(60/800)*(1/1000) ]

I multiplied with 1/2 because probability of selecting any group at random from 2 group is 1/2.

I hope this is clear now.

wonderful explanation sachin. thanks.

(60/1000)*(1/800)=((60/800)*(1/1000)=3*40000

Please notice that, the question is "will be a sibling pair", so, answer is not: (60/1000)*(60/800)=9/2000.

Why is it not 60/800 as well?

If there 60/1000 in the junior class, wouldn't a corresponding set of 60 be in the senior class, hence 60/800? Why is it 1/800?

Thank you!

sudi760mba wrote:Why is it not 60/800 as well?

If there 60/1000 in the junior class, wouldn't a corresponding set of 60 be in the senior class, hence 60/800? Why is it 1/800?

Thank you!

Once a student is chosen from the Junior class, there will be a unique student in the Senior class, as a sibling for the chosen one.

We're really choosing the students simultaneously, so in our calculations it makes no difference which one is selected "first".

We could certainly calculate it two different ways:

60/1000 * 1/800

OR

60/800 * 1/1000

However, these aren't two distinct possibilities (in which case we'd add the results together), these are the same possibility expressed two different ways (so we don't add them together).

Here's another way to look at it that may be clearer:

probability = (# of desired outcomes) / (total # of possibilities)

There are 60 matching pairs we could select, so we have 60 desired pairs.

We're selecting 1 person from each subgroup, so there are 800C1 * 1000C1 (or simply 800*1000) total possible pairs.

Therefore, Prob(matching pair) = 60/(800*1000)

Stuart Kovinsky wrote:We're really choosing the students simultaneously, so in our calculations it makes no difference which one is selected "first".

We could certainly calculate it two different ways:

60/1000 * 1/800

OR

60/800 * 1/1000

However, these aren't two distinct possibilities (in which case we'd add the results together), these are the same possibility expressed two different ways (so we don't add them together).

Here's another way to look at it that may be clearer:

probability = (# of desired outcomes) / (total # of possibilities)

There are 60 matching pairs we could select, so we have 60 desired pairs.

We're selecting 1 person from each subgroup, so there are 800C1 * 1000C1 (or simply 800*1000) total possible pairs.

Therefore, Prob(matching pair) = 60/(800*1000)

Hi Stuart,

I followed your explanation until " There are 60 matching pairs we could select, so we have 60 desired pairs."
Then you lost me, unfortunately. I understand that the desired outcome is 60 matches, which is comprised of 60 picks from juniors and 60 picks from seniors. Isn't that a combined total # of possibilities of (60/1000)*(60/800)?

Let's assume I don't learn the logic behind this probability before my exam in 3 weeks...What would be a logical way to eliminate answer choices?