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Mathematics 19 Online
OpenStudy (anonymous):

Pat picked a card from a standard deck, looked at it, and then put it back. He then picked a second card. What is the probability that Pat picked a diamond or a jack on either pick? sixteen over fifty two one over fifty two thirteen over fifty two seventeen over fifty two

OpenStudy (anonymous):

what is the probability he got in on the first pick?

OpenStudy (kirbykirby):

If he picked a card then replaced it into the deck, you have the same deck again in the 2nd pick. So figure out the probability of picking a diamond, \(P(D)\), the probability of picking a jack, \(P(J)\) Then, the probability of picking a diamond OR jack (in one single pick), then \(P(D \text{ or } J ) = P(D \cup J) = P(D) + P(J) - P(D \cap J)\) Now to get the probability of "diamond or jack" on either turn, means that you could get: "Diamond or Jack" on pick 1, OR "diamond or Jack" on pick 2. So: turn 1: the probability is \(P(D \cup J)\) turn 2: the probability is \(P(D \cup J)\) The overall probability is then: \(P(D \text{ or } J \text{ on either turn})=P(D \cup J)+P(D \cup J)\).. there is no intersection to worry about for this one because the 1st pick and 2nd pick are mutually exclusive

OpenStudy (kirbykirby):

If you want to be explicit about the last statement I wrote^^, well you could theoretically write it in a more cumbersome way, following the "regular" rule for unions as: \( P\left((D \text{ or } J)_{\text{turn 1 } }\cup (D \text{ or } J)_{\text{turn 2 } }\right)\\=P(D \cup J)_{\text{turn 1 } }+P(D \cup J)_{\text{turn 2 } }-P\left((D \text{ or } J)_{\text{turn 1 } }\cap (D \text{ or } J)_{\text{turn 2 } }\right)\) But as I said, the intersection of the event \((D \text{ or } J)_{\text{turn 1 } }\cap (D \text{ or } J)_{\text{turn 2 } }\) is 0 since both turns cannot occur simultaneously.

OpenStudy (anonymous):

ok soo im thinking its A

OpenStudy (kirbykirby):

Ah hm yes I think you are right. According to the choices, adding P(D or J) + P(D or J) doesn't work. I guess I misinterpreted "either" as being "inclusive or", when I think it should rather mean "exclusive or", so the probability should just be P(D or J) which is the same as answer A

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