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2 years ago

A four-person committee is chosen from a group of eight boys and six girls.

Mathematics

2 answers:

Stolb23 [73]2 years ago

6 0

Answer:

The probability is approximately 0.0699. That's $\displaystyle \frac{10}{143}$ to be exact.

Step-by-step explanation:

How many ways to choose four out of a total of $8 + 6 = 14$ boys and girls?

$\displaystyle {14\choose 4} = 1001$ .

How many ways to choose four boys out of a total eight boys?

$\displaystyle {8\choose 4} = 70$ .

What's the probability that all four persons chosen out of the pack of fourteen are boys? All 1001 ways of choosing four out of fourteen are equally likely. However, only 70 satisfy the conditions. In other words, the probability of choosing four boys out of this group of eight boys and six girls is equal to:

$\begin{aligned}&\frac{\text{Number of ways to choose four out of eight}}{\text{Number of ways to choose four out of fourteen}} \\= &\frac{70}{1001}\\=& \frac{10}{143}\approx 0.0699\end{aligned}$ .

Viefleur [7K]2 years ago

5 0

Answer:

10/143

Step-by-step explanation:

got 100%

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A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$