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3 years ago

Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?

Mathematics

1 answer:

photoshop1234 [79]3 years ago

4 0

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

$D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6$

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

$T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326$

What is the probability that they are both queens?

$P = \frac{D}{T} = \frac{6}{1326} = 0.0045$

0.45% probability that they are both queens.

Read more about products at:

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3 0

1 year ago

Please help me out solve it and tell me how you got that answer so I can understand please!

NikAS [45]

The solution for s in the given equation is $s = C-nD$

The question seems to be incomplete

Here is the complete question:

Solve for s in this equation $D= \frac{C-s}{n}$ Depreciation.

To solve for s, that means we should make s the subject of the equation

From the given equation,

$D= \frac{C-s}{n}$

To solve for s, first multiply both sides by n to clear the fraction

We get

$n\times D= n \times \frac{C-s}{n}$

Then,

$nD = C - s$

Now, add s to both sides

$nD + s= C - s+s$

$nD+s = C$

Then, subtract nD from both sides

$nD-nD+s = C-nD$

∴ $s = C-nD$

Hence, the solution for s in the given equation is $s = C-nD$

Learn more here: brainly.com/question/21406377

6 0

2 years ago

I have 30 photos to post on my website. I'm planning to post these on two web pages, one marked "Friends" and the other marked "

Sergio039 [100]

Answer:a) 870

b) 435

Step-by-step explanation:

number of photos to be posted =n = 30

number of web pages on which they would be posted is 2

Since the order in which the photos appear on the web pages matters,

Number of ways = 30 permutation 2

=870 ways

Since the order in which the photos appear on the web pages does not matter,

Number of ways = 30 combination 2

= 435 ways

6 0

3 years ago

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5-6}{0.2}$

$Z = -5$

$Z = -5$ has a pvalue of 0.

1 - 0 = 1

5 0

3 years ago