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Probability (m01q04) [#permalink]
 06 May 2008, 04:57
Question Stats:
79% (01:33) correct
20% (00:39) wrong based on 1 sessions
Three cannons are firing at a target. If their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively, what is the probability that none of the cannons will hit the target after one round of fire? (A) 0.06 (B) 0.12 (C) 0.21 (D) 0.29 (E) 0.94 Source: GMAT Club Tests - hardest GMAT questions Probability of getting it right : 3/10 * 4 /10 * 5 /10 = 3/50 Not getting any right = 1 - 3/50 = 47/50 = .94 Please confirm if this is the correct answer. Thanks
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alimad wrote: Three cannons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively. What is the probability that none of the salvos will hit the target? (Salvo is one round of fire.)
Probability of getting it right : 3/10 * 4 /10 * 5 /10 = 3/50
Not getting any right = 1 - 3/50 = 47/50 = .94
Please confirm if this is the correct answer. Thanks I think it will be 0.7*0.6*0.5 = 0.21
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I concur.
Alimad, you found out the probab of -
Getting none right + getting 1 right + getting 2 right
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prob of none hitting them = (1-prob of cannon a hitting)*(1-prob of cannon b hitting)*(1-prob of cannon c hitting)
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pmenon wrote: prob of none hitting them = (1-prob of cannon a hitting)*(1-prob of cannon b hitting)*(1-prob of cannon c hitting) I agree with you: (1-0.3)*(1-0.4)*(1-0.5)=0.21 Regards [Edited the multiplication]
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Last edited by JohnLewis1980 on 08 May 2008, 05:51, edited 1 time in total.
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alimad wrote: Three cannons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively. What is the probability that none of the salvos will hit the target? (Salvo is one round of fire.)
Probability of getting it right : 3/10 * 4 /10 * 5 /10 = 3/50
Not getting any right = 1 - 3/50 = 47/50 = .94
Please confirm if this is the correct answer. Thanks Should be 7/10*6/10*1/2 ---> 21/100 or 21%
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alimad wrote: Three cannons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively. What is the probability that none of the salvos will hit the target? (Salvo is one round of fire.)
probability to hit is given for each, so for each we can find the probability to miss and multiply those together. (1-0.3)(1-0.4)(1-0.5) = (0.7)(0.6)(0.5) = 0.21
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When you multiple all of the chances of hitting the target together, that .06 is actually the chance of getting 3 hits (1 from each cannon) in 1 round of fire. Obviously, with those probabilities given (.3, .4, & .5) that isn't very high. So think of it as the inverse, "What is the chance that all 3 will miss?" That would be .7 chance of a miss * .6 chance of a miss * .5 chance of a miss = 0.21. So, how would you compute the probability of getting at least 1 hit from the 3 shots? Would that then be 79% chance? Because if there is a 21% chance all will miss, and we're looking for anyhing but that, it would be 1 - .21, for .79? That makes sense, what if the question were asking about at least 2 hits in 1 round?
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alimad wrote: Three cannons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively. What is the probability that none of the salvos will hit the target? (Salvo is one round of fire.)
Probability of getting it right : 3/10 * 4 /10 * 5 /10 = 3/50
Not getting any right = 1 - 3/50 = 47/50 = .94
Please confirm if this is the correct answer. Thanks you found the probability that at-least one canon misses the shot..... right answer = (1-0.3)(1-0.4)(1-0.5) = 0.21
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Re: Probability (m01q04) [#permalink]
30 Aug 2009, 10:54
Probability of first canon hitting = P(A) = 0.3
Probability of second canon hitting = P(B) = 0.4
Probability of third canon hitting = P(C) = 0.5
hence Probability of first canon not hitting = 1 - P(A) = 0.7
hence Probability of second canon not hitting = 1 - P(B) = 0.6
hence Probability of third canon not hitting = 1 - P(C) = 0.5
Total Probability that none of the canons hit
= (1 - P(A)) * (1 - P(B)) * (1 - P(C))
= 0.7 * 0.6 * 0.5
= 0.21
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Re: Probability (m01q04) [#permalink]
29 Jan 2010, 19:43
BEsides you can eliminate some answer choice by strategic guessing by looking at the answer choices: For instance the probability of cannon A hitting the target - 0.30 The probability of B hitting the target is - 0.40 the probability of C hitting the target is - 0.50 ( and his probability of not hitting the target is also 0.50) thus the answer definitely CANNOT be 0.94 ( eliminate choice E) Answer choice 0.06 is too low thus eliminate as well. So you ;'re left with B, C and D. Well, this works if you're pressed under the time. OF course, it's better to be 100% sure though CHeerz
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Re: Probability (m01q04) [#permalink]
10 Oct 2010, 05:58
"what is the probability that none of the cannons will hit the target after one round of fire?" The question has confused me.
I took it this way: What is the probability that in any round after the first, none of the cannons will hit the target.
I guess the question implies that after the first round is over and we then check which cannon has hit the target, the answer should be none. In this case the question should use "will have" instead of "will"...
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Re: Probability (m01q04) [#permalink]
31 Oct 2010, 02:52
alimad wrote: Three cannons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively. What is the probability that none of the salvos will hit the target? (Salvo is one round of fire.)
Probability of getting it right : 3/10 * 4 /10 * 5 /10 = 3/50
Not getting any right = 1 - 3/50 = 47/50 = .94
Please confirm if this is the correct answer. Thanks I want to explain why your approach is incorrect 3/10 * 4/10 * 5/10 = 3/50 is the probability of getting three right. You did not count the probability of getting one right, two rights.
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Re: Probability (m01q04) [#permalink]
25 Nov 2010, 11:52
1st cannons will not hit the target after one round of fire will be=(1st hit * 2nd doesn’t hit) OR (1st doesn’t hit * 2nd doesn’t hit)=.3 * .7 + .7*.7 =.21+.49 =.7 With the same reasoning we get for 2nd and 3rd canons
= .6 and .5So finally none of the target will hit after one shots of fire will be multiplication of all these possibilities = .7*.6*.5 =.21 It's much easier if you make a tree diagram. I did one but don’t know how to put that from word to this posting 
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Re: Probability (m01q04) [#permalink]
25 Nov 2010, 19:43
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tanvirkhan77 wrote: 1st cannons will not hit the target after one round of fire will be=(1st hit * 2nd doesn’t hit) OR (1st doesn’t hit * 2nd doesn’t hit)=.3 * .7 + .7*.7 =.21+.49 =.7 With the same reasoning we get for 2nd and 3rd canons
= .6 and .5So finally none of the target will hit after one shots of fire will be multiplication of all these possibilities = .7*.6*.5 =.21 It's much easier if you make a tree diagram. I did one but don’t know how to put that from word to this posting whoa whoa whoa there!! Dont get so worked up buddy. All we have to do is multiply the individual probabilities of all 3 missing the target since 1 must miss AND 2 must miss AND 3 must miss. 0.7*0.6*0.5 = 0.21. (In the situations where we end up in an OR situation, we add individual probabilities). Though we all have different things clicking for us, there is such a thing as "faster way".
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Re: Probability (m01q04) [#permalink]
25 Nov 2010, 22:53
i agree with 0.21. its will be the multiple of chances that will not hit the target. so, it will be (1-0.3)*(1-0.4)*(1-0.5) =0.21
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Re: Probability (m01q04) [#permalink]
27 Nov 2010, 07:43
C. Prob of not hitting of each cannon seperately is .7, .6, .5 if all of them do not hit the individual prob need to be multiplied. These are simultaneous events ( not mutually exclusive ) p = .7*.6*.5=.21 Posted from my mobile device 
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Re: Probability (m01q04) [#permalink]
30 Nov 2010, 13:58
alimad wrote: Three cannons are firing at a target. If their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively, what is the probability that none of the cannons will hit the target after one round of fire? (A) 0.06 (B) 0.12 (C) 0.21 (D) 0.29 (E) 0.94 Source: GMAT Club Tests - hardest GMAT questions Probability of getting it right : 3/10 * 4 /10 * 5 /10 = 3/50 Not getting any right = 1 - 3/50 = 47/50 = .94 Please confirm if this is the correct answer. Thanks
I did (7/10)*(3/10)*(5/10) = 210/1000
Move the decimal 3 places to the left and you get .210. Answer C
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Re: Probability (m01q04) [#permalink]
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 
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Re: Probability (m01q04) [#permalink]
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)
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10 Jan 2011, 09:11
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