Question

In a batch of 8,000 clock radios 5% are defective. A sample of 14 clock radios is randomly selected without replacement from the 8,000 and tested. The entire batch will be rejected if at least one of those tested is defective. What is the probability that the entire batch will be rejected?
A) 0.488
B) 0.0714
C) 0.512
D) 0.0500

Answer 1

Answer:

0.512

Step-by-step explanation:

Answer 2

1. $\frac{5}{100} *8000= 5*80=400$ of radios are defective, so           8000-400=7600 work properly.

2. P(entire batch rejected)=P(at least one out of randomly selected 14 is defecive)= 1-P(none of the randomly selected 14 is defective)

3. Lets make the previous point clear. The complement or the opposite of "at least one defective" is "none defective" or "all working".

4. so lets find the probability of 14 being all working:

P(all 14 randomly selected, working)=$\frac{7600}{8000}* \frac{7599}{7999}* \frac{7598}{7998}* \frac{7597}{7997}*...* \frac{7588}{7988}* \frac{7587}{7987}$

$\frac{7600}{8000}=0.95$
$\frac{7599}{7999}=0.95$
.
.
.
$\frac{7587}{7987}=0.95$

So it is clear that each of these 14 fractions is very close to 0.95,

P(all 14 randomly selected, working)=0.95*0.95...0.95=$(0.95)^{14}= 0.488$

(to calculate easily, in your computers calculator, write 0.95, hit *, then hit Enter 13 times)

P(none of 14 selected working)=1-P(all working)=1-0.488=0.512

4. Another solution to P(all working) is $\frac{C(7600, 14)}{8000}= \frac{ \frac{7600!}{14!(7600-14)!} }{8000}= \frac{ \frac{7600!}{14!(7586)!} }{8000}=\frac{ \frac{7600*7599*7598*...*7587*7586!}{14!(7586)!} }{8000}$

$=\frac{ \frac{7600*7599*7598*...*7587}{14!} }{8000}$

here we need to multiply all the numbers from 7600 to 7587 and then divide by 14, by 13.....1. This result then is divided by 8000, and what we get, is substracted from 1.

5. The principles discussed, of at least one of the solutions, should be clear at this level of problem.