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Problem: I have a box that contains balls: 2 black and 5 white. I have the right to take 2 balls in a row without returning. Find the probability that two chosen balls are white: Since these events are independent:

$$P(A) = P(A_1) + P(A_2) = \frac{5}{7} + \frac{4}{6} = \frac{29}{21}>1$$ -> why?

Localth
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John Doe
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    Whenever you get a strange result you should re-evaluate your assumptions. – John Douma Sep 10 '20 at 16:40
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    You need to multiply, not add. – Nuclear Hoagie Sep 10 '20 at 16:41
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    $P(A_1 \cap A_2)=P(A_2|A_1)P(A_1)$ –  Sep 10 '20 at 16:44
  • Nuclear Wang but we multiply when the events occur at the same time. These events don't occur at the same time so why we should multiply – John Doe Sep 10 '20 at 16:45
  • independent events multiply themselves, not add. – luisfelipe18 Sep 10 '20 at 16:45
  • when do we add then? – John Doe Sep 10 '20 at 16:48
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    Probabilities add for disjoint events, multiply for independent events. – Sandejo Sep 10 '20 at 16:55
  • I suspect someone has been explaining probability to you rather casually. As Luis Felipe says, you add probabilities when you're talking about the probability of any of a number of disjoint events. For example, the probability that a family has no more than two children is equal to the probability that they have no children, plus the probability that they have one child, plus the probability they have two children. – Brian Tung Sep 10 '20 at 17:15
  • We multiply when we want the probability of all of a number of independent events. For instance, the probability that it rains today and my coin flip comes up heads is equal to the probability that it rains today, times the probability that my coin flips heads. – Brian Tung Sep 10 '20 at 17:16
  • In this case, the multiplication is of a different sort (at least intuitively), specified by Matthew Holder in their comment: The probability that two things happen—independent or not—can be expressed as the probability the first thing happens, times the conditional probability that the second thing happens given that the first thing happens. So, for example, suppose I hit a baseball with probability $1/4$. Given that I hit it, the probability that it is caught "on the fly" is $1/3$. Then the probability that I hit a ball that is caught on the fly is $1/4 \times 1/3 = 1/12$. – Brian Tung Sep 10 '20 at 17:20
  • You can think of it like this: Out of any $12$ times, I'll hit the baseball $3$ times (on average), and of those $3$ times, it is caught on the fly $1$ time (on average). – Brian Tung Sep 10 '20 at 17:20

2 Answers2

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For your example to verify the correct answer consider listing all of the possible and equally likely arrangements of the 2 black and 5 white balls:

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One can count that out of the 21 possible arrangements, 10 of those have the first two balls chosen being white.

That should convince you that adding the two probabilities is not appropriate.

JimB
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The two events are dependent events. You need the first one (getting first ball to be white) "and" the second event to happen ( getting the second ball to be white) at the same time . So it is "and" relation between two events to happen (be True) at the same time. ( "And" relation i.e. both events to be true not "Or" relation i.e. either one of the two events is only true ) So you have to multiply the two events' probabilities not adding

Shereen Ali
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