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 Post subject: probability question from veritas prep exam 4Posted: Sat Sep 25, 2010 10:59 pm

Joined: Thu Jul 08, 2010 11:09 pm
Posts: 33
the sets s1= (15, 9, 20,7) and s2 =(6,11,5,4) both have one or more numbers over 10

If one number from S1 and one number from s2 are selected randomly, what is the probability that the set selected will have one number over 10 and one number under 10?

a) 3/16
b) 3/8
c) 7/16
d) 1/2
e) 5/8

ans: D

i get that 4*4 would be the total possible outcomes, but i do not understand how to arrive at the possible outcomes. Is it using a combination or permutation. The explanation in the exam was too vague and unclear and actually seems wrong. Please help me understand how to arrive at the numerator value thanks.

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 Post subject: Re: probability question from veritas prep exam 4Posted: Sun Sep 26, 2010 2:12 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
I'd actually go about this one a slightly different way - and maybe this will make more sense to you.

There are two ways we can achieve our desired outcome.

A. We pick a number OVER 10 from S1 and a number UNDER 1 from S2.
B. We pick a number UNDER 10 from S1 and a number OVER 1 from S2.

Let's calculate the probability of each of these and then add them, since it's a probability of either A OR B.

A.
Probability of picking number>10 from S1 is 1/2
Probability of picking number <10 from S2 is 3/4
Multiply these and we get 3/8
B.
Probability of picking number <10 from S1 is 1/2
Probability of picking number >10 from S2 is 1/4
Multiply these and we get 1/8

Now, add 3/8 and 1/8 together and we get 4/8, or 1/2.

Veritas Help

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 Post subject: Re: probability question from veritas prep exam 4Posted: Mon Sep 27, 2010 2:04 pm

Joined: Thu Jul 08, 2010 11:09 pm
Posts: 33
ok that makes a lot more sense to me now
thanks.

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