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2 years ago

Find the probability of each event. Answer IN FRACTION form.

Mathematics

1 answer:

Sloan [31]2 years ago

4 0

Total amount of students = 6

Total amount of girls = 4

So, chance of picking a girl first time is 4/6 or 2/3 chance

But remember! The girl is picked, and we do not want to replace her back into the set. So total amount of students drop to 5 and girls drop to 3

So using that logic, The probability of choosing all girls = 4/6 * 3/5 * 2/4 * 1/3 = 1/15.

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You have two summer jobs. In the first job, you work 25 hours per week and earn $7.75 per hour. In the second job, you earn $6.2

Morgarella [4.7K]

Answer:

3.4 Hours

Step-by-step explanation:

25 x $7.75 = $193.75

$250 - $193.75 = $21.25

$21.25/$6.25 = 3.4 hours

You must work 3.4 hours at the second job.

3 0

2 years ago

Read 2 more answers

Does 1/2+1/8 = to 3/4

AleksAgata [21]

1/2 is 2/4, and so we would need another 1/4 to make it equal to 3/4
1/8 is not equal to 1/4, so 1/2+1/8 does not equal 3/4

5 0

2 years ago

Read 2 more answers

How much more area does a large pizza with a 12in. diameter have than a small pizza with a 8 in. diameter? Round your answer to

USPshnik [31]

Answer:

A is the correct answer.

Step-by-step explanation:

First do 6 times 6 and then do 36 times 3.14 to get 113.04.

Then do 4 times 4 to get 16 times 3.14 for 50.24

Last do 113 minus 50 for 63 squared.

6 0

2 years ago

Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

2 years ago

Tim picked 1 7/12 buckets of apples, and Fred picked 4 2/3 buckets of apples. How many more buckets did Fred pick?

ElenaW [278]

The answer is 2 buckets

Explain:
Tim picked 1 7/12 and Fred picked 4 2/3
These as improper fractions are 19/12
And 8/3
To make these have the same denominator you would times the 8/3 by 4 to give 32/12
Take these two away, Fred has 13 more apples.
And as we calculated it with there being 12 in each bucket there would be one full bucket but we would need another bucket for the apple left
So Fred picked 2 more buckets than Tim.
Hoped it makes sense and that I helped

3 0

2 years ago