What is the Probability of Drawing Certain Numbers from a Bag of Counters?

[Mo]

I absolutely hate stats! its got to be one of the worst topics going.Anyway here is the (simple) question i am having (lots) of problems with! (im revising over and can't remember a thing!)

"In each of the following questions, list a sample space consisting of equally likely outcomes and hence find the required probabilities"

1) There are 5 counters in a bag, each marked with a different number from the set {1,2,3,4,5} .Two counters are drawn from the bag,Find the probabaility the total score on the counters is

a) greater than 7
b) a prime number*
c) an odd number*

I made up a sample space diagram and i find out that out of 20 possible outcomes, 4 added up to be greater than 7. So my answer was 1/5

However i reread it and i done it through another way ... i said:

p(grater than 7) is the probabaility we get... p(3,5) + p(5,3) + p(4,5) + p(5,4)

(ok so this is essentially the same) BUT the probability of getting a 3 is 1/5 right? so we get (1/5 X 1/5) X 4 = 4/25

so ... which one i correct?

Please, please help!

Regrads
Mo


* i think ill attempt these again once i know the answer for this.

 

[Galileo]

I made up a sample space diagram and i find out that out of 20 possible outcomes, 4 added up to be greater than 7. So my answer was 1/5

That's correct. It was unnecessary to consider (3,5) different from (5,3) though. There are ${5 \choose 2} = 10$ ways of picking two counters and only (3,5) and (4,5) give a sum greater than 7. So the prob. is 1/5.
Same answer and both methods are valid.

p(grater than 7) is the probabaility we get... p(3,5) + p(5,3) + p(4,5) + p(5,4)

(ok so this is essentially the same) BUT the probability of getting a 3 is 1/5 right? so we get (1/5 X 1/5) X 4 = 4/25

Remember that once you have taken a counter (say 3) out of the bag there are 4 left. So the probability of picking the 5 after you've taken the 3 is not 1/5, but 1/4.

 

[Mo]

thank you Galileo!

Regards,
Relieved - Mo