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Probability (m01q04)

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Re: Probability (m01q04) [#permalink] New post 02 Dec 2010, 03:24
Probability
Cannon A hits: 0.3
Cannon A doesn't hit: 0.7
Cannon B hits: 0.4
Cannon B doesn't hit: 0.6
Cannon C hits: 0.5
Cannon C doesn't hit: 0.5

The probability that none of the cannons will hit the target after one round:
(A doesn't hit) x (B doesn't hit) x (C doesn't hit) = 0.7 x 0.6 x 0.5 = 0.21

The probability that at least one cannon will hit the target after one round:
1 - (A, B, and C all don't hit) = 1 - (none of the cannons hits) = 1 - 0.21 = 0.79

The probability that at least two cannons will hit the target after one round:
(A doesn't hit, B hits, C hits) + (A hits, B doesn't hit, C hits) + (A hits, B hits, C doesn't hit) + (A hits, B hits, C hits)
= (A doesn't hit) x (B hits) x (C hits) + (A hits) x (B doesn't hit) x (C hits) + (A hits) x (B hits) x (C doesn't hit) + (A hits) x (B hits) x (C hits)
= 0.7 x 0.4 x 0.5 + 0.3 x 0.6 x 0.5 + 0.3 x 0.4 x 0.5 + 0.3 x 0.4 x 0.5
= 0.35

The probability that all three cannons will hit the target after one round:
(A hits) x (B hits) x (C hits) = 0.3 x 0.4 x 0.5 = 0.06
8-)
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Re: Probability (m01q04) [#permalink] New post 10 Jan 2011, 09:11
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.
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Re: Probability (m01q04) [#permalink] New post 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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Re: Probability (m01q04) [#permalink] New post 01 Dec 2011, 01:02
are there any good books from where one can learn probability?
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Re: Probability (m01q04) [#permalink] New post 05 Dec 2011, 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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Re: Probability (m01q04) [#permalink] New post 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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Re: Probability (m01q04) [#permalink] New post 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
Re: Probability (m01q04)   [#permalink] 29 Nov 2012, 12:15
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