Probabilty questions....these are really bugging me?

Question

ok...
"A quality control inspector examines 5 parts from a bin of 25. 6 parts are defective. What is hte probabilty that no more than 1 defective part will be examined?
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I can only get 25C5 so far, and I'm fairly sure that goes on the bottom of a division. I'm just not sure what goes on top.

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If there is a 20% chance of rain for each of the next seven days, what is the probability it will rain at least one of the next seven days?
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I know that the answer to this one is 79%, I just can't figure out how to get it.

~thanks in advance

Answer 1

Okay, you're right about the 25C5, as there are that many ways to choose five parts out of 25. No more than 1 defective part chosen means that either zero or one defective part was chosen.

PROBABILITY THAT ZERO DEFECTIVE PARTS CHOSEN:
There are 19 functioning parts and so 19 C 5 ways to choose five functioning parts.
(19 C 5) / (25 C 5)

PROBABILITY THAT ONE DEFECTIVE PART IS CHOSEN:
There are 19 functioning parts and four such must be chosen. Further, there are 6 defective part of which one must be chosen.
(19 C 4) (6 C 1) / (25 C 5)

SO THE PROBABILITY THAT EITHER ZERO OR ONE PARTS ARE DEFECTIVE IS:
[(19 C 5) + (19 C 4) (6 C 1)] / (25 C 5)

I don't have a calcator with me, so you're going to have to figure it out.

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Anytime you see "atleast" in a probability question, it should be a red flag. Sure, you could calculate the probability that it rains one day, or two days, or three days, or four days, etc, etc and add it all together, but making use of the complement is much easier.

Atleast one means anything but zero, in other words:
1 - P(zero days rain)

We know that each day has a 0.80 probability not raining and we assume the events are independent. So the probability of it not raining all seven days is (0.80)^7.

Thus, probability it rains at least one day:
1 - (0.80)^7 = 0.79

Answer 2

Question one:

There are two possibilities: (A) He finds no defective parts OR (B) He finds one defective part. This is the same as the event where he finds no more than one defective object.

Event A:
What is the probability he finds no defective parts?
He must've chosen 5 objects from the 19 non-defective objects. Thus: (19C5)/(25C5)

Event B:
What is the probability he finds one defective part? He chooses 4 objects from the 19 non-defective objects and then chooses 1 from the 6 defective objects.
Thus: (19C4)*(6C1)/(25C5)

Since P(A or B) = P(A) + P(B) - P(A and B), we know that P(A or B) = P(A) + P(B) since P(A and B) = 0 because the two events are mutually exclusive. Add these two probabilities together to get the probability that no more than one defective part will be examined.




Question 2:

We need to know the probability that it will rain on at least one of the next seven days. This is the same as the complement of it NOT raining at all during the next seven days. So, let's find the probability that it won't rain during the next seven days.

What's the probability it won't rain on any one day?
We know there's a 20% chance of rain, so there's an 80% chance of it not raining on any one day.

The probability of it not raining 7 days in a row is just (0.80)^7 = 21%.

Thus, the probability of it raining at least 1 day in the next 7 is just 100%-21% = 79%.


Let me know if anything wasn't clear.

Answer 3

"A quality control inspector examines 5 parts from a bin of 25. 6 parts are defective. What is the probability that no more than 1 defective part will be examined?"

Exact approach:
Probability
= [C(19,5)+C(6,1)C(19,4)]
/C(25,5)
= 0.6566

Since the sample of 25 is very large compared to the 5 picks. You can use binomial distribution to approximate the result. You pick zero or one defective part out of five picks.

probability
= C(5,1)(6/25)(19/25)^4
+C(5,0)(19/25)^5
= 0.6539
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"If there is a 20% chance of rain for each of the next seven days, what is the probability it will rain at least one of the next seven days?"
1-(1-0.2)^7 = 79%

Answer 4

1st problem
You're right, 25C5 is the bottom. On top, number of ways of picking NO defective parts is 19C5, and number of ways of picking 1 defective part is 6C1 • 19C4, so your prob is

19C5 + 6C1 • 19C4
-------------------------- =
25C5

11628 + 6(3876)
------------------------ =
53130

3876/8855 = 0.4377

2nd
prob rain at least 1 day is 1 - prob rain NO days,
1 - 0.8^7 = 1 - 0.2097 = 0.7903

Answer 5

i think 6c5 goes on the top of the first one but i cant figure out the second one sorry

Answer 6

find that none are defective
19nCr5/25nCr5
then 1-answer

2. find none
1-80^7=