<PRE><B>Question 177396</B>
A can contains 7 pink balls and 5 blue balls. 
FOUR balls are selected without replacement. 
What is the probability of 
1) P(all pink or all blue) = (7C4 + 5C4)/12C4 
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2) P(exactly 3 pink) = 7C3*5C1/12C4
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3) P(at least 3 pink) = P(3 pink) + P(4 pink) = [7C3*5C1]/12C4 + 5C4/12C4
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4) P(exactly 3 pink or 3 blue) = P(3 pink) + P(3 blue)
=  7C3*5C1/12C4 + 5C3*7C1/12C4
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Cheers,
Stan H.</PRE>