Ask question

Ask question

All categories

3 years ago

10

A chef is choosing from 13 different entrees for an important banquet, 3 of which contain spinach. The guests will have a choice

of 4 entrees. The chef has decided to choose the 4 at random. Write your answers as a decimals rounded to the nearest thousandth. What is the probability that none of the chosen entrees contain spinach

1 answer:

lina2011 [118]3 years ago

4 0

Answer:

0.294 = 29.4% probability that none of the chosen entrees contain spinach.

Step-by-step explanation:

The choices of entrees are chosen without replacement, which means that the hypergeomtric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

13 different entrees for an important banquet

This means that $N = 13$

3 of which contain spinach.

This means that $k = 3$

The guests will have a choice of 4 entrees.

This means that $n = 4$

What is the probability that none of the chosen entrees contain spinach?

This is P(X = 0). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,13,4,3) = \frac{C_{3,0}*C_{10,4}}{C_{13,4}} = 0.294$

0.294 = 29.4% probability that none of the chosen entrees contain spinach.

You might be interested in

Find the sum of the arithmetic sequence 22, 11, 0, -11, ..., –77.

blagie [28]

Answer:

B.

Step-by-step explanation:

First you need to figure out how many numbers there are, so with this you can simply count 10. Then use the formula to plug in the numbers: 10(22-77)/2 to get the answer.

7 0

4 years ago

A rectangle has a length of 42 inches and a width of 14 inches. Which of the following is not equal to the ratio of the rectangl

Temka [501]

The ratio of the rectangle's length to its width is 42/14 which simplifies to 3/1 by factoring out 14 out of the numerator and denominator.
Any ratio that is not 3:1, or a multiple of 3:1, is not equal.

List of Equal Ratios: 6:2, 9:3, 12:4, etc.

4 0

3 years ago

Is 49 a perfect square yes or no

Romashka [77]

Answer:

Yes

Step-by-step explanation:

7 x 7 = 49

4 0

3 years ago

Read 2 more answers

dan spent half of his allowance going to the movies he washed the family car and earn $10 what is his weekly allowance if you en

Angelina_Jolie [31]

12 to 14 is his weekly allowance

7 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

4 years ago