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Probability (m01q04) : Retired Discussions [Locked]

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Probability (m01q04) [#permalink]
06 May 2008, 03:57

2

This post was BOOKMARKED

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?

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? _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: Probability (m01q04) [#permalink]
25 Nov 2010, 18:43

1

This post received KUDOS

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 .5

So 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". _________________

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

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

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.

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

Re: Probability (m01q04) [#permalink]
30 Aug 2009, 09:54

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

Re: Probability (m01q04) [#permalink]
29 Jan 2010, 18:43

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

Re: Probability (m01q04) [#permalink]
10 Oct 2010, 04:58

"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"...

Re: Probability (m01q04) [#permalink]
31 Oct 2010, 01:52

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.

Re: Probability (m01q04) [#permalink]
25 Nov 2010, 10:52

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 .5

So 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

Re: Probability (m01q04) [#permalink]
27 Nov 2010, 06:43

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

Re: Probability (m01q04) [#permalink]
30 Nov 2010, 12:58

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?

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

If the individual probability of each cannon hitting their target is 3/10, 4/10, and 5/10 that would mean the chance of each cannon MISSING is 7/10, 6/10, and 5/10 respectively.

I did (7/10)*(3/10)*(5/10) = 210/1000

Move the decimal 3 places to the left and you get .210. Answer C _________________

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!

gmatclubot

Re: Probability (m01q04)
[#permalink]
30 Nov 2010, 12:58