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quester [9]
3 years ago
5

We are drawing two cards without replacement from a standard​ 52-card deck. find the probability that we draw at least one red c

ardred card.
Mathematics
1 answer:
alina1380 [7]3 years ago
8 0
The probability of drawing only one red card in two draws without replacement is given by:
P(1\ red)=\frac{26C1\times26C1}{52C2}=0.51
The probability of drawing two red cards in two draws without replacement isgiven by:
P(2\ red)=\frac{26C2\times26C0}{52C2}=0.245
The events 'draw one red card' and 'draw two red cards' are mutually exclusive. Therefore the probability of drawing at least one red card is 0.51 + 0.245 = 0.755.
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<span> On the first day of Christmas,
my true love sent to me
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The song continues, adding 4 calling birds on the 4th day, 5 golden rings on the 5th, and so on up to the 12th day, when 12 drummers add to the cacophony of assorted birds, pipers and lords leaping all over the place.

Notice that on each day there is one partridge (so I will have 12 partridges by the 12th day), and each day from the second day onwards there are 2 doves (so I will have 22 doves), and from the 3rd there are 3 hens (total of 30 hens), and so on.

So, how many presents are there altogether?

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Doves: 2 × 11 = 22

Hens 3 × 10 = 30

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Total = 364

We observe that we have the same number of partridges as drummers (12 of each); doves and pipers (22 of each); hens and lords (30 of each) and so on. So the easiest way to count our presents is to add up to the middle of the list and then double the result: (12 + 22 + 30 + 36 + 40 + 42) × 2 = 364.


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3 years ago
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