greed1 wrote:

Hi I was wondering if anyone knows how to work this question out:

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

all the balls chosen are different colours?

2 of the balls are the same colour?

3 of the balls are the same colour?

all the balls are the same colour?

Working out the all same and all different is not too hard but is there a quick way to do the other parts without drawing the 625 branch tree diagram?

1) All balls of different colors --> (5/5) * (4/5) * (3/5) * (2/5) = (24/125)

2) 2 balls of the same color --> ((pick the color to be repeated twice) * (pick the other 2 color) * (number of permutations))/(Total permutations) = (5 * C(4,2) * (4!/2!))/(5^4) = (72/125)

3) 3 balls of the same color --> ((pick the common color) * (pick the other color) * (number of permutations))/(total permutations) = (5 * C(4,1) * (4!/3!))/5^4 = (16/125)

4) All same color --> Only 1 such combination possible for each color, there are 5 colors, so total probability = 5/5^4 = (1/125)

5) Two balls of one color and two of another color --> ((choose 2 colors) * (permute))/(total permutations) = C(5,2) * (4!/2!2!) / 5^4 = (12/125)Total probability for all cases = 1 _________________

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1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

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