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Probability (m01q04)

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Re: Probability (m01q04) [#permalink] New post 10 Jan 2011, 08:22
petrifiedbutstanding wrote:
Can someone please tell me where I can find some good coaching on probability? An online course would be the best. If that isn't available, I'll settle for whatever is good.


LOL.. I would like to request the same thing....
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Re: Probability (m01q04) [#permalink] New post 01 Dec 2011, 00:02
are there any good books from where one can learn probability?
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Re: Probability (m01q04) [#permalink] New post 05 Dec 2011, 01:04
probability of not hitting the target = pro(1st cannon not hitting)*pro(2nd not hitting)*pro(3rd not hitting)
= .7*.6*.5
= .21
hence C
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Re: Probability (m01q04) [#permalink] New post 29 Nov 2012, 05:30
If A = the event that the 1st cannon hits the target
B = the event that the 2nd cannon hits the target
C = the event that the 3rd cannon hits the target

We need to find out P(A' * B' * C') where A' = the event that 1st cannon does not hit the target etc and '*' comes as we need the event A' AND B' AND C'.

As A, B, C are independent events, so is A', B', C' ; hence, P(A'*B'*C') = P(A')*P(B')*P(C') = (1-P(A))*(1-P(B))*(1-P(C)) = 0.7*0.6*0.5 = 0.21 (Ans)

So the correct choice is 'C'.
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Re: Probability (m01q04) [#permalink] New post 29 Nov 2012, 11:15
Getting right is given for each canon. Then getting wrong is 1- probability of getting right for each canon. Probability of happening each case is AND operation whereas either case is OR. If AND then multiply all the probabilities.
In this case probability of getting all wrong P is,
(1-0.3)*(1-0.4)*(1-0.5) =0.21
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Re: Probability (m01q04) [#permalink] New post 12 Mar 2013, 03:50
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0.3, 0.4, and 0.5

This one is easy----

The probability of the guns not firing = .7,.6, .5

So the probability of gun 1 not firing AND gun 2 not firing AND gun 3 not firing = .7 *.6* .5= .21
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Re: Probability (m01q04) [#permalink] New post 12 Mar 2013, 04:55
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can anyone post a good source for prob questions
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Re: Probability (m01q04) [#permalink] New post 19 Sep 2013, 03:42
I have seen so many of you telling the answer correctly. And kudos to you! I have to expose myself for thinking lamely. So, no one, of the participating pundits, seemed to have made a mistake of doing this 1- (0.3*0.5*0.6)=.94. What did I do? Or in other words, when would I do what I did, if ever? Yet, in other words, how should the problem sound so as to make my way of solving correct? I am no longer consider myself bad at probability, but this problem definitely prompts further training...:)

Thanks in advance.
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Re: Probability (m01q04) [#permalink] New post 19 Sep 2013, 03:57
0.06 is the probability that all three canons hit the target. The other 0.94 includes the probability that none, one, and two cannons hit the target. Correct me if I am wrong.
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Re: Probability (m01q04) [#permalink] New post 19 Sep 2013, 04:01
Expert's post
Juz2play wrote:
0.06 is the probability that all three canons hit the target. The other 0.94 includes the probability that none, one, OR two cannons hit the target. Correct me if I am wrong.


Should be OR, otherwise it's correct.
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Re: Probability (m01q04) [#permalink] New post 04 Dec 2013, 10:44
Probability of A hitting the target = 0.3 (probability of miss = 1 - 0.3 = 0.7)
Probability of B hitting the target = 0.4 (prob of miss = 0.6)
Probability of C hitting/missing the target = 0.5

Possible outcomes = 8 ( MMM, MMH, MHM, MHH, HMM, HMH, HHM, HHH)

Easiest way to calculate the missing probability is: directly multiply MMM = 0.7 X 0.6 X 0.5 = 0.21

Answer is C
Re: Probability (m01q04)   [#permalink] 04 Dec 2013, 10:44
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