Probability Question?

Question

In a shipment of 15 room air conditioners, there are 3 with defective thermostats. Two air conditioners will be selected at random and checked one after another. Find the probability that:

a. the first is defective
b. the first is defective and the second is good
c. both are defective
d. the second one is defective
e. exactly one of the 2 is defective
f. at least one is defective

Thanks for your help :)

Answer 1

alright. so there are 15 possible choices for the first pick, and 14 possible choices for the second pick. (once you pick one, you have one less to pick another from)

so for part a. out of the 15 air conditioners, 3 of them are defective. If you choose one air conditioner out of the 15, you have a 3/15 chance of choosing a defective one. so the probability of choosing a defective one on the first try is 3/15, which is 20%.

for part b, once you choose the first one to be defective, there are only 2 defective ones left and 12 good ones. So if you assume that the first is defective, then the probability of choosing the second one to be good is 12/14. This is only if the first one is defective though. So you know the probability of the first being defective. You just multiply the 2 values, 12/14 * 3/15 = (6/7) * (1/5) = 6/35 = 17.14%

for part c, you do basically the same thing as in b, but instead of the second one being good, it is defective. so the probability is 2/14 instead of 12/14. so the probability is (2/14) * (3/15) = (1/7)*(1/5) = 1/35 = 2.86%

for part d, this one is a bit trickier. alright. so you found the probability of the second one being defective when you know the first one is defective (answer in part c). but what if the first one was not defective? Say the first one was not defective. The probability of this happening is 12/15 (12 good ones out of 15 possible choices). Now, if you want the second to be defective, then out of 14 possible choices, there are 3 that are defective. The probability of choosing a defective one is (3/14). So you multiply these 2 probabilities: (3/14)*(12/15) = (3/14) * (4/5) = 12/70 = 6/35
Now, you have the probability of choosing a defective second one when the first is defective (1/35) and you have the probability of choosing a defective second one when the first is not defective (6/35). So the total probability of choosing a defective second one is the sum of these 2 probabilities (7/35) = 20%
If you think about this, then you will realize that you basically found the probability of the second one being defective no matter what the first one is.
When they ask you a question like this, then you have to go through every single case and find the probability of what u want in each case. So in this question, you went through every possible case of the first choice and found the probability of choosing a defective second one for each choice of the first one. and then you added the probabilities.

part e.
alright, so if exactly 1 of the 2 is defective, then either the first one is defective and second is good or the second is defective and first is good.
so you have to find the probability of both of these cases and add it up.
the first case: the first is defective and second is good. This is the answer you found in part b. This is 6/35. Now, if you want the second to be defective and first good, you found this in d. This is also 6/35. This is just a coincidence, it does not always happen this way.
Anyways, so the total probability of exactly one of them being defective is the sum of those 2, which is 12/35 = 34.29%

f.
so now they say at least one is defective. This means that either exactly one is defective, or both of them are defective. So the way you would do this is to first find the probability that exactly one is defective (done in part e), and add it to the probability that both of them are defective (foudn in part c).
So the answer is (12/35) + (1/35) = 13/35 = 37.14%
The probability of at least one being defective is 37.14%.

I hope you understood all that.

Answer 2

a. There's 15 total so you would set up a fraction. There are 3 chances of getting a defective one out of 15, so you get 3/15 or 1/5.

b. There's 15 total and 3/15 are defective and then there's going to be 14 left. 12 out of 14 are not defective, so you get 6/7. You would then multiply 6/7 times 1/5 and you get 6/35.

c. There's 15 total and 3/15 [1/5] are defective. In order to get the other one defective is 2/14 because one is gone and so you have to subtract one from each. Then you get 1/5 times 1/7 and you get 1/35.

d. There's 15 total and 1/5 are defective so you already have that one and there's 12 out of 14 that are normal. Therefore you get the same answer as b.

e. You would make a chart with 1-15 on the top and 1-14 on the bottom and you see how many times you get the number 1 2 or 3 with another number other than 1, 2, or 3 on it.

f. You would do the same chart and count how many times you get a 1, 2, or a 3 by itself or with another number.

Hope I helped.

Answer 3

a) there are 3 defective ones out of 15. Probability is 3/15 or 1/5.

b) after you select one air conditioners, there will be 14 left. The the probability of getting a good one is 12/14.
So probability is 1/5 * 12/14 = 6/35

c) 3/15 * 2/14 = 1/35

d) you only select one air conditioner, If it is the second one, what about the first one? defective or good?

e) exactly one of the two.
the favor combinations is 3*12=36
posisible combinations is 105

36/105 = 12/35

f) 12*3 = 36
(3*2)/2 = 3

36+3 = 39
probability is 39/105 or 13/35

Answer 4

3/15
3/15*12/14=6/35
3/15*2/14 = 1/35
12/15* 3/14 = 6/35
3/15*12/14*2 = 12/35

Answer 5

So, in 15 air conditioners, 3 are bad and 12 are good.

a. 3/15 = 20%
b. (3/15)(12/14) = 17.14%
c. (3/15)(2/14) = 2.857%
d. (3/15)(2/14) + (12/15)(3/14) = 20%
e. (3/15)(12/14) + (12/15)(3/14) = 34.29%
f. 1 - (12/15)(11/14) = 37.14%

Answer 6

a. 1/5
b. (1/5)(6/7)
c. (1/5)(1/7)
d. 1/5
e. (1/5)(6/7) + (4/5)(3/14)
f. 1-(4/5)(11/14)

Answer 7

a) 20%
b) 17.1%
c) 2.8%
d) 24.2%
e) 34.2%
f) 37%

I could be wrong i'd wait for back up. i was quite good at probability though!