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Two balls are selected at random from a bag with three red balls and two yellow balls. Find…

a) probability that the first ball is red

3/5 . 2/4 + 3/5 . 2/4 = 3/5

b) probability that the second ball is red

2/5 . 3/4 + 3/5 . 2/4 = 3/5

c) probability that both are red

(3c2)/(5c2) = 3/10

d) probability that the second ball is red, given that the first ball is red

P(A|B) = P(AB)/P(B) = (3/10)/(3/5) = 1/2

Are my workings correct?

Parcly Taxel
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MasterYoshi
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    Yes, they are all correct, though your arithmetic in parts (a) and (b) are unnecessary. It was also unnecessary to approach directly via definition of conditional probability for part (d). – JMoravitz Jul 20 '18 at 03:07
  • unnecessary in what way? I only know how to do it that way or lets say for a) [(3c1)(2c1)+(3c1)(2c1)]/ 20 . (20 permutations) – MasterYoshi Jul 20 '18 at 03:09
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    A reminder: when selecting a sample space for an experiment, one need not necessarily keep track of all information. You only need to keep track of the information that is relevant to you to solve the problem and select your sample space in such a way that each outcome described is equally likely to occur. For part (a), we label the balls and we only need to keep track of what the first ball drawn is (and completely ignore the second ball), giving only five outcomes in our sample space, each of which equally likely to occur, and only three of which refer to a red ball first. – JMoravitz Jul 20 '18 at 03:10
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    To the unseasoned student, many would approach (b) the same way you did, thinking that the fact that it is the second ball we are talking about that this is somehow relevant. That's not the case as can be seen in related questions like this one. You could have just treated your sample space the same way as for (a), containing only five outcomes but here referring to the outcome of the second ball only (ignoring the first). – JMoravitz Jul 20 '18 at 03:14
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    For (d), rephrase it away from a conditional probability question to a direct question: "What is the probability that you draw a red ball from a bag with two red balls and two yellow balls?" After all, this would be the scenario you would be in if it were given that you drew a red ball in the first round. – JMoravitz Jul 20 '18 at 03:15
  • I really appreciate the explanation. I get too caught up in just plugging equation's in , bad habit. Thank you, I feel enlightened! – MasterYoshi Jul 20 '18 at 03:18

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