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Eva8 [605]
3 years ago
10

A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (i)a card of s

pade or an ace (ii)a red king (iii)neither a king nor a queen (iv)either a king or queen.
Thank you.
Mathematics
1 answer:
docker41 [41]3 years ago
5 0
There are 52 cards in a standard deck so that would be the whole in the part/whole equation so
i... There are 13 spade cards and then 3 other aces since there are 4 suits of cards so 16/52 or about 0.30769 or 30.8% rounded

ii....There are 2 reds of each card in a standard deck along with 2 blacks so there would be 2 red kings which would be a 2/52 chance or about 0.0384 or 3.8% rounded

iii.... There are 4 kings and 4 queens in each deck so (52-8)/52 which is a 44/52 chance, 0.8461 about, or 84.6% rounded

iv..., As I said in iii, there are 8 kings and queens so an 8/52 chance, about 0.1538, or 15.4% rounded
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We would have to take a sample of 62 to achieve this result.

Step-by-step explanation:

Confidence level of 95%.

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

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That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg.

This means that \sigma = 2

If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result?

We would need a sample of n.

n is found when M = 0.5. So

M = z\frac{\sigma}{\sqrt{n}}

0.5 = 1.96\frac{2}{\sqrt{n}}

0.5\sqrt{n} = 2*1.96

Dividing both sides by 0.5

\sqrt{n} = 4*1.96

(\sqrt{n})^2 = (4*1.96)^2

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Rounding up

We would have to take a sample of 62 to achieve this result.

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Answer:

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We still have to pay for 10 minutes.

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Hope this helps!

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