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# Probability (m01q04)

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Manager
Status: Trying to get into the illustrious 700 club!
Joined: 18 Oct 2010
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Kudos [?]: 23 [0], given: 58

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10 Jan 2011, 09:22
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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I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

Manager
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01 Dec 2011, 01:02
are there any good books from where one can learn probability?
Manager
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05 Dec 2011, 02:04
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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-------Analyze why option A in SC wrong-------

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Joined: 04 Aug 2012
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Concentration: General Management, Operations
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29 Nov 2012, 06:30
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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29 Nov 2012, 12:15
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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12 Mar 2013, 04:50
2
KUDOS
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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12 Mar 2013, 05:55
2
KUDOS
can anyone post a good source for prob questions
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19 Sep 2013, 04:42
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...

Manager
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19 Sep 2013, 04:57
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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Don't be afraid to fail, but be afraid not to try

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19 Sep 2013, 05:01
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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04 Dec 2013, 11:44
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

Re: Probability (m01q04)   [#permalink] 04 Dec 2013, 11:44

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# Probability (m01q04)

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