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
what is the probability he got in on the first pick?
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
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.
ok soo im thinking its A
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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