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Probability Problem

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by alsergi » Wed Aug 20, 2008 5:33 am
Please, help me with this problem ...


A couple on holiday has made a list of 10 cities it wishes to visit, four of which are in Northern Europe and six of which are in Southern Europe. As the couple has only time to visit three of these ten cities, it chooses three of the ten cities at random. What is the probability that the couple will visit at least one city in Northern Europe and one city in Southern Europe?

Answers:

a.-1/2
b.-2/3
c.-3/4
d.-4/5
e.-5/6

Thanks!!!
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by parallel_chase » Wed Aug 20, 2008 6:07 am
10 cities

4 North
6 South

3 cities in total

1 at least N, 1 at least S.

2N * 1S = (4C2 * 6C1)/10C3 = (6 * 6)/120

1N * 2S = (4C1 * 6C2)/10C3 = (4 * 15)/120

(6 * 6)/120 + (4 * 15)/120 = 4/5


Hope this helps.
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by pepeprepa » Wed Aug 20, 2008 6:38 am
Here is my way:

P(X): what we are looking for
P(N): probability we only have north cities
P(S): probability we only have south cities

P(X)=1-P(N)-P(S)

Total possibilities of the 3 choices of cities is: 10C3=120

Total possibilities of having only 3 cities of north: 4C3=4
P(N)=4/120

Total possibilities of having only 3 cities of south: 6C3=20
P(S)=20/120

Finally,
P(X)=1-4/120-20/120=96/120=4/5

so D
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