SOLUTION: A card is drawn from a full deck of 52 cards. What is the probability that the card is either a queen or a red card?
--I got 4/52 that a queen is drawn, and 26/52 that a red ca
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-> SOLUTION: A card is drawn from a full deck of 52 cards. What is the probability that the card is either a queen or a red card?
--I got 4/52 that a queen is drawn, and 26/52 that a red ca
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Question 1088806: A card is drawn from a full deck of 52 cards. What is the probability that the card is either a queen or a red card?
--I got 4/52 that a queen is drawn, and 26/52 that a red card is drawn. Would the final answer be 30/52? or am I completely lost.... Answer by jim_thompson5910(35256) (Show Source):
P(Q) = probability that you draw a queen
P(Q) = (number of queens)/(number of cards total)
P(Q) = 4/52
P(Q) = 1/13
P(R) = probability that you draw a red card
P(Q) = (number of red cards)/(number of cards total)
P(R) = 26/52
P(R) = 1/2
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P(Q and R) = probability you pick a queen AND a red card
P(Q and R) = probability you pick a red queen
P(Q and R) = P(Q)*P(R) ... works because Q and R are independent events
P(Q and R) = (1/13)*(1/2)
P(Q and R) = (1*1)/(13*2)
P(Q and R) = 1/26
Note: An alternative is to write it out like this
P(Q and R) = (number of red queens)/(number of cards total)
P(Q and R) = 2/52
P(Q and R) = 1/26
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P(Q or R) = probability of queen OR a red card (or both, i.e. a red queen)
P(Q or R) = P(Q) + P(R) - P(Q and R) see note below
P(Q or R) = (1/13) + (1/2) - (1/26)
P(Q or R) = (2/26) + (13/26) - (1/26)
P(Q or R) = (2+13-1)/26
P(Q or R) = 14/26
P(Q or R) = 7/13
Note: You are adding correctly (when you add 4/52 to 26/52); however, you forgot to subtract off the value 1/26. Why is this subtraction done? To correct for double counting. There is overlap between Q and R: namely the 2 red queens. We count each of those red queens twice when we compute P(Q)+P(R). This is why we subtract off P(Q and R) to correct for this overcounting.
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