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An urn contains 5 red, 6 blue and 8 green balls. 3 balls are randomly selected from the urn, find the probability that they are all of the same color if:

(a) the balls are drawn without replacement;

(b) the balls are drawn with replacement.

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What have you tried? –  Thomas Andrews Feb 4 '13 at 21:22
    
what have you tryed? Hint: P(all 3) = P(drawing one of color a)*P(drawing another of color a)*P(drawing yet another one of color a) –  kaine Feb 4 '13 at 21:22

1 Answer 1

$1.$ With replacement: There are three ways this can happen, red, red, red; blue, blue, blue; and green, green, green.

The probability of red, red, red is $\left(\dfrac{5}{19}\right)^3$. Find similar expressions for the other two colurs, and add up.

$2.$ Without Replacement: The probability of red, red, red is $\left(\dfrac{5}{19}\right)\left(\dfrac{4}{18}\right)\left(\dfrac{3}{17}\right)$. Find similar expressions for the other two colours, and add up.

Remark: We could also do the problem by a counting argument. Let's look at the first problem. Put labels on the balls to make them distinct. Then for with replacement, there are $19^3$ strings of length $3$ of our objects. (The three objects in the string are not necessarily distinct.) All of these $19^3$ strings are equally likely.

There are $5^3$ all red strings, $11^3$ all blue, and $8^3$ all green. This gives probability $\dfrac{5^3+6^3+8^3}{19^3}$.

One can do a similar calculation for the without replacement case.

For without replacement, there is a third approach. We can choose $3$ objects from $19$ in $\dbinom{19}{3}$ ways. And we can choose $3$ of the same colour in $\dbinom{5}{3}+\dbinom{6}{3}+\dbinom{8}{3}$ ways.

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