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## Probability Question - People Who Rock At Math, Please Take A Look?

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### Probability Question - People Who Rock At Math, Please Take A Look?

There are 10 essays, lets call them a,b,c,d,e,f,g,h,i, and j.

They are in pairs, i.e 'a' and 'b' together, 'c' and 'd' together and so on, making 5 pairs.

The teacher says she will ask 3 essays on the exam, of which we can write any 1.

However, she will only ask 1 essay from a pair. There is no possibility of her asking both the essays from the same pair.

If I skip pairs 'cd' and 'ij', which leaves me with 3 pairs of essays, what is the probability of essays from those two skipped pairs coming in the exam?

(I tried to explain this as simply as i could, sorry if it's not clear.)

Thank you!
caspar

Posts: 481
Joined: Thu Mar 31, 2011 8:57 pm

### Probability Question - People Who Rock At Math, Please Take A Look?

what you mean by skip? like you don't have to do them?
Raimondo

Posts: 46
Joined: Mon Jan 13, 2014 1:11 pm

### Probability Question - People Who Rock At Math, Please Take A Look?

I believe it would be 2/5×3/5 which is 6/25 or 24%
Flann

Posts: 40
Joined: Sun Feb 16, 2014 7:04 pm

### Probability Question - People Who Rock At Math, Please Take A Look?

the teacher will always give only 1 essay from a pair, and you are also skipping pairs, so we can deal with each pair as 1 entity

P(at least 1 essay from skipped pairs)

= 1 - P(all from unskipped pairs)

= 1 - 3c3/5c3 = 1 - 1/10 = 0.9
Richman

Posts: 43
Joined: Wed Jan 01, 2014 2:24 am

### Probability Question - People Who Rock At Math, Please Take A Look?

For your question, as long as you study both halves of any pair,

then you no longer have to treat them as pairs.

Let's call the pairs A C E G J [ because I looks like 1 'eye' looks like 'one' ]

If you study A E G only,

since only 3 pairs will appear on the exam,

the worst case will be that the 3 on the exam

are { C J and (one you studied) }.

But since you only have to write one of them, you're ok.

Just ignore C and J, and concentrate on doing the other one really well.

So what are the probabilities of the various scenarios?

There are 10 possible sets of 3 essays the teacher can pick. (5c3)

Let x = one of A, E, G

C J x = 3 cases (one for each of x = A, E, G)

C x x = 3 cases (omit ONE of A E G, include the other two)

J x x = 3 cases (same as previous)

A E G = 1 case (no choices here)

P( you studied all 3 that appear on the exam ) = 1/10 = (3c3) / (5c3)

P( you studied 2 out of the 3 on the exam ) = 3/10 = (3c2) / (5c3)

P( you studied 1 out of the 3 on the exam ) = 6/10 = (2*3c1) / (5c3)

P( you studied 0 that appear on the exam ) = 0

(i.e. you're guaranteed to have studied at least one)

or

P( both CJ ) = 3/10

P( one of them ) = 6/10

P ( neither ) = 1/10

P(at least one of them) = 9/10
Edur

Posts: 49
Joined: Wed Feb 05, 2014 10:06 pm