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There are 3 red balls and 2 blue balls in a box.
a) Draw a probability tree for 2 balls selected randomly from the box if the selection is done with replacement.
b) Draw a probability tree for 2 balls selected randomly from the box if the selection is without replacement.
c) If 3 balls are selected without replacement, what is the probability that the first ball is red, the second and the third balls are blue?
b) red, red
red, blue
blue, red
blue, blue
c) the chance of first ball red is 1/5 and second ball blue is 1/2 and third ball blue is 1/3.
I have a different answer for C) The probability that the first ball is red is 3/5, the second ball blue is 6/20 and the third ball blue is 6/60. Professor please clarify this question and tell us who has the correct answer, if any of us.
The chance that the first ball is 3/5. With only 4 balls left (2r,2b), the chance that the next is blue is 2/4=1/2 and finally, with 3 balls left (2r, 1b), the chance that the 3rd ball is blue is 1/3. Multiplying these probabilities gives 3/5*1/2*1/3=1/10 (cancel the 3’s!)
For parts a) and b), you need to draw trees. I have attached a hand-drawn result. (Please note that the book does trees differently, but I will accept if you follow their model.)
Whenever I was drawing the trees, I found that for some reason my number were not matching the Probability of 1, that is, that to verify your answer, to some extent, all of the “leaves” in turn must add to a total of 1. I found my error after looking at the tree from the professor.
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