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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 01:53
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Three of the 12 cars at the car dealership were flooded cars. If someone purchased two cars from the dealership at random, what is the probablility that neither of the cars were flooded cars?



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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 02:02
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P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 02:47
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ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22
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ywilfred:,

didn't you calculate the probability of both cars being flooded. The question asks for both not flooded.

Ans : 1-1/22 = 21/22?
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 02:59
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ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22
Show more


I think you calculated for both cars flooded.

Prob of first car not flooded = P1 = 9/12
Prob of second car not flooded = P2 = 8/11

Prob of both cars not flooded = P1 * P2 = 72/132 = 6/11
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 03:10
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ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22

I think you calculated for both cars flooded.

Prob of first car not flooded = P1 = 9/12
Prob of second car not flooded = P2 = 8/11

Prob of both cars not flooded = P1 * P2 = 72/132 = 6/11
Show more


Dahiya,

I think in your solution, you should also calculate 2 more cases where:

1) First car is flooded, second not flooded = 3/12*9/11
2) First car not flooded, second flooded = 9/12*3/11

So P(Neither flooded) = 6/11 (Your answer) + 9/44 + 9/44 = 21/22
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 03:25
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ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22

I think you calculated for both cars flooded.

Prob of first car not flooded = P1 = 9/12
Prob of second car not flooded = P2 = 8/11

Prob of both cars not flooded = P1 * P2 = 72/132 = 6/11

Dahiya,

I think in your solution, you should also calculate 2 more cases where:

1) First car is flooded, second not flooded = 3/12*9/11
2) First car not flooded, second flooded = 9/12*3/11

So P(Neither flooded) = 6/11 (Your answer) + 9/44 + 9/44 = 21/22
Show more


If I am not wrong, "what is the probablility that neither of the cars were flooded cars" means none of them is flooded.

We have four cases

1. F F = (3/12) * (2/11) = 1/22
2. F N = (3/12) * (9/11) = 9/44
3. N F = (9/12) * (3/11) = 9/44
4. N N = (9/12) * (8/11) = 6/11 This is the question asked.

6/11 + 9/44 + 9/44 = 21/22 is answer when asked "Atleast one of them is not flooded"

Where I am going wrong?? :roll: :roll: :roll: :roll:
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 03:37
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ps_dahiya

1. F F = (3/12) * (2/11) = 1/22
2. F N = (3/12) * (9/11) = 9/44
3. N F = (9/12) * (3/11) = 9/44
4. N N = (9/12) * (8/11) = 6/11 This is the question asked.

6/11 + 9/44 + 9/44 = 21/22 is answer when asked "Atleast one of them is not flooded"

Where I am going wrong?? :roll: :roll: :roll: :roll:
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You are absolutely right Sir! Sorry about confusing you besides being confused.

I should probably go to bed now
:pc
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 05:03
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joemama142000
Three of the 12 cars at the car dealership were flooded cars. If someone purchased two cars from the dealership at random, what is the probablility that neither of the cars were flooded cars?
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What is wrong with my reasoning?

prob of choosing the first flooded=3/12
prob of choosing the second flooded=2/11

prob of choosing both flooded= 3/12* 2/11=6/132=1/22

prob that neither flooded =1-(1/22)? =21/22

ahhh, i suck at probability

the oa is 9C2/12C2

does that make sense?
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 05:34
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2 out of 12 cars can be selected in 66 ways. Remove the 3 flooded cars, 9 left .You need to select 2 out of 9 in 36 ways. The required prob is 36/66 or 2C9/2C12
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 10:19
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joemama142000
joemama142000
Three of the 12 cars at the car dealership were flooded cars. If someone purchased two cars from the dealership at random, what is the probablility that neither of the cars were flooded cars?

What is wrong with my reasoning?

prob of choosing the first flooded=3/12
prob of choosing the second flooded=2/11

prob of choosing both flooded= 3/12* 2/11=6/132=1/22

prob that neither flooded =1-(1/22)? =21/22

ahhh, i suck at probability

the oa is 9C2/12C2

does that make sense?
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Yes it is the same value as 6/11 when calculated. It should be 6/11 as explained previously.
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 13:37
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i would also like to know where joemamas logic is incorrect, since it is also what i followed.

3/12 cars are flooded. Question asks what is probability that out of two cars selected, NEITHER of them are flooded ?

P(neither flooded) = 1- prob(both flooded) = 1 -[(3/12)*(2/11)] = 1-1/22 = 21/22

Can someone explain ?
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 13:55
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pmenon
i would also like to know where joemamas logic is incorrect, since it is also what i followed.

3/12 cars are flooded. Question asks what is probability that out of two cars selected, NEITHER of them are flooded ?

P(neither flooded) = 1- prob(both flooded) = 1 -[(3/12)*(2/11)] = 1-1/22 = 21/22

Can someone explain ?
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P = 1- prob(both flooded) is the probability when at least one of them is not flooded. See my previous explanation.

P(neither flooded) = 1 - P(Exactly one flooded) - P(Both flooded)
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 19:18
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giddi77
ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22

ywilfred:,

didn't you calculate the probability of both cars being flooded. The question asks for both not flooded.

Ans : 1-1/22 = 21/22?
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oops..haha.. :oops: It should be 1-1/22 = 21/22
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 22:26
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Even i got it wrong :oops:

I got 9/ 16

P(Flooded)=3/12=1/4
so P(Non F)= 1-1/4=3/4
I just did 3/4 * 3/4 to get 9/16

Why did this logic didnt work???
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ywilfred
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ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22

ywilfred:,

didn't you calculate the probability of both cars being flooded. The question asks for both not flooded.

Ans : 1-1/22 = 21/22?


oops..haha.. :oops: It should be 1-1/22 = 21/22
Show more


ywilfred,

Either I am confused or you are??

How it could be 21/22 instead of 6/11??
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Three of the 12 cars at the car dealership were flooded cars. If some 09 Jan 2006, 22:54
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ywilfred
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ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22

ywilfred:,

didn't you calculate the probability of both cars being flooded. The question asks for both not flooded.

Ans : 1-1/22 = 21/22?


oops..haha.. :oops: It should be 1-1/22 = 21/22

ywilfred,

Either I am confused or you are??

How it could be 21/22 instead of 6/11??
Show more


What I did was to subtract the probability that both cars are flooded from 1. This is wrong since that will leave me with sets that have 1 car flooded. Sorry that this fact slipped my mind. The neat way to do it would be this way:

P(car not flooded) and P(car not flooded) = 9/12 * 8/11 = 3/4 * 8/11 = 6/11

You can choose to do the way that you did, which was to use combinations. That is, 9C2/12C2 = 6/11.

Yet, you can do it the long way if you like by:

P(car flooded) and P(car flooded) = 3/12 * 2/11 = 1/4 * 2/11 = 1/22
P(car flooded) and P(car not flooded) = 3/12 * 9/11 = 1/4 * 9/11 = 9/44
P(car not flooded) and P(car flooded) = 9/12 * 3/11 = 3/4 * 3/11 = 9/44

P(both cars not flooded) = 1-1/22-9/44-9/44 = 6/11
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ps_dahiya
ywilfred
P(not flooded)
= P(first car not flooded) AND P(second car not flooded)
= 3/12 * 2/11 = 1/4 * 2/11
= 2/44 = 1/22

I think you calculated for both cars flooded.

Prob of first car not flooded = P1 = 9/12
Prob of second car not flooded = P2 = 8/11

Prob of both cars not flooded = P1 * P2 = 72/132 = 6/11
Show more



YES !!...I agree with ur solution...I calculated the same way
6/11..
wats the OA !!??
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vikasgaba
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its 6/11. question says prob of not flooded. so its shall be (9/12)*(8/11) = 6/11



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