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Probabilitry

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Probabilitry

by Md Raihan Uddin » Thu Oct 02, 2014 12:51 pm
Three canons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4 and 0.5. What is the probability that none of the canons would hit the target?
1)0.06
2)0.12
3)0.21
4)0.29
5)0.94
First Method
= 1-(0.3*0.4*0.5)
=0.94
2nd Method
(1-0.3)*(1-0.4)*(1-0.5)=0.7*0.6*0.5=0.21

Now my question is why 1st method is wrong and why 2nd one is correct. How will I understand which one to use and when?
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by GMATGuruNY » Thu Oct 02, 2014 1:18 pm
Md Raihan Uddin wrote: First Method
= 1-(0.3*0.4*0.5)
=0.94

Now my question is why 1st method is wrong
The portion in red represents the probability that ALL 3 cannons hit the target.
Thus:
1 - (0.3 * 0.4 * 0.5) = 0.94 = the probability that NOT ALL 3 CANNONS hit the target.

There are 3 ways that NOT ALL 3 CANNONS may hit the target:
Case 1: No cannon hits the target.
Case 2: Exactly 1 cannon hits the target.
Case 3: exactly 2 cannons hit the target.

Thus:
0.94 = P(no cannon hits the target OR exactly 1 cannon hits the target OR exactly 2 cannons hit the target).
Not what the question is asking.
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by Brent@GMATPrepNow » Thu Oct 02, 2014 2:52 pm
Md Raihan Uddin wrote:Three canons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4 and 0.5. What is the probability that none of the canons would hit the target?
A) 0.06
B) 0.12
C) 0.21
D) 0.29
E) 0.94
P(1st MISSES target) = 1 - P(1st HITS target) = 1 - 0.3 = 0.7
P(2nd MISSES target) = 1 - P(2nd HITS target) = 1 - 0.4 = 0.6
P(3rd MISSES target) = 1 - P(3rd HITS target) = 1 - 0.5 = 0.5
----------------------------

P(none hit the target) = P(1st misses target AND 2nd misses target AND 3rd misses target)
= P(1st misses target) x P(2nd misses target) x P(3rd misses target)
= 0.7 x 0.6 x 0.5
[spoiler]= 0.21
= C[/spoiler]

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by [email protected] » Thu Oct 02, 2014 10:46 pm
Hi Md Raihan Uddin,

In these types of multi-part probability questions, the details matter a great deal, so you should force yourself to take plenty of notes. In that way, you'll know exactly what each "math step" will be. My guess is that your math skills are just fine, so the issue is really about how you organize your work.

Here, we're told the individual probabilities for three canons to hit a target (.3, .4 and .5, respectively). We're asked for the probability that NONE of the canons hit the target.

From a note-taking standpoint, here is one way that you could write down the question:

(Prob that 1st misses)(Prob that 2nd misses)(Prob that 3rd misses) = ?

From here, you can just input the number and multiply....

(.7)(.6)(.5) = .21

Final Answer: C

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by pooja181 » Thu Oct 02, 2014 10:56 pm
The portion in red represents the probability that ALL 3 cannons hit the target.
Thus:
1 - (0.3 * 0.4 * 0.5) = 0.94 = the probability that NOT ALL 3 CANNONS hit the target.

There are 3 ways that NOT ALL 3 CANNONS may hit the target:
Case 1: No cannon hits the target.
Case 2: Exactly 1 cannon hits the target.
Case 3: exactly 2 cannons hit the target.

Thus:
0.94 = P(no cannon hits the target OR exactly 1 cannon hits the target OR exactly 2 cannons hit the target).
Not what the question is asking.
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by Matt@VeritasPrep » Fri Oct 03, 2014 12:11 pm
It often helps to get all your probabilities in order first.

First Cannon:
Hit = .3
Miss = .7

Second Cannon:
Hit = .4
Miss = .6

Third Cannon:
Hit = .5
Miss = .5

If we want ALL THREE cannons to hit, we'd have .3 * .4 * .5.

If we do 1 - (.3 * .4 * .5), we'd have the chances that NOT ALL THREE cannons hit; in other words, the odds that AT LEAST ONE cannon misses.

The important thing here is that the negation of "all" is not "none", it's "not all"! If I say "not all the cannons hit", some of them could still hit the target; there's just (at least) one that misses. This logic will come into play in critical reasoning too, so it's worth practicing.
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