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# M01-04

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Math Expert
Joined: 02 Sep 2009
Posts: 47983
M01-04  [#permalink]

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16 Sep 2014, 00:14
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Difficulty:

15% (low)

Question Stats:

79% (00:33) correct 21% (00:50) wrong based on 140 sessions

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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

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Posts: 47983
Re M01-04  [#permalink]

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16 Sep 2014, 00:14
Official Solution:

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

If the probability of an event is $$P(A)$$, the probability that the event will not happen is $$1 - P(A)$$. If there are two independent events: $$A$$ and $$B$$, the probability that they both occur equals the product of the probabilities of the two individual events:
$$P(C) = P(A) * P(B)$$

Using these two theorems:
$$P = P (not\ H_1) * P(not\ H_2) * P(not\ H_3)$$
$$P = (1-0.3)(1-0.4)(1-0.5)=$$
$$= 0.7*0.6*0.5=0.21$$

Where $$H$$ represents the cannon hitting the target after one round of fire.

Answer: C
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Joined: 04 Sep 2015
Posts: 36
Location: Germany
Concentration: Operations, Finance
WE: Project Management (Aerospace and Defense)
Re: M01-04  [#permalink]

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06 Jan 2016, 14:23
Bunuel wrote:
Official Solution:

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

If the probability of an event is $$P(A)$$, the probability that the event will not happen is $$1 - P(A)$$. If there are two independent events: $$A$$ and $$B$$, the probability that they both occur equals the product of the probabilities of the two individual events:
$$P(C) = P(A) * P(B)$$

Using these two theorems:
$$P = P (not\ H_1) * P(not\ H_2) * P(not\ H_3)$$
$$P = (1-0.3)(1-0.4)(1-0.5)=$$
$$= 0.7*0.6*0.5=0.21$$

Where $$H$$ represents the cannon hitting the target after one round of fire.

Answer: C

HI Bunuel

Can you please let me know what's wrong in below approach

P [None] = 1 - P [all hit target]
= 1 - (0.3) (0.4) (0.5)
= 0.94

Thanks!
Manager
Joined: 19 Dec 2015
Posts: 111
Location: United States
GMAT 1: 720 Q50 V38
GPA: 3.8
WE: Information Technology (Computer Software)
Re: M01-04  [#permalink]

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07 Jan 2016, 21:30
The probability of 0.94 ( As calculated by you ) represents all scenarios except all canons hitting the target at the same time. So this can include, 1 canon hitting the target, 2 hitting the target and "None hitting the target". You have to find just "None hitting the target". And thus this approach is incorrect in this question.

MK1480 wrote:
Bunuel wrote:
Official Solution:

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

If the probability of an event is $$P(A)$$, the probability that the event will not happen is $$1 - P(A)$$. If there are two independent events: $$A$$ and $$B$$, the probability that they both occur equals the product of the probabilities of the two individual events:
$$P(C) = P(A) * P(B)$$

Using these two theorems:
$$P = P (not\ H_1) * P(not\ H_2) * P(not\ H_3)$$
$$P = (1-0.3)(1-0.4)(1-0.5)=$$
$$= 0.7*0.6*0.5=0.21$$

Where $$H$$ represents the cannon hitting the target after one round of fire.

Answer: C

HI Bunuel

Can you please let me know what's wrong in below approach

P [None] = 1 - P [all hit target]
= 1 - (0.3) (0.4) (0.5)
= 0.94

Thanks!
Intern
Joined: 04 Sep 2015
Posts: 36
Location: Germany
Concentration: Operations, Finance
WE: Project Management (Aerospace and Defense)
Re: M01-04  [#permalink]

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08 Jan 2016, 16:24
Thanks for your clarification!

[quote="trehan"]The probability of 0.94 ( As calculated by you ) represents all scenarios except all canons hitting the target at the same time. So this can include, 1 canon hitting the target, 2 hitting the target and "None hitting the target". You have to find just "None hitting the target". And thus this approach is incorrect in this question.
Senior Manager
Joined: 06 Jan 2015
Posts: 444
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
Re: M01-04  [#permalink]

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27 Jan 2018, 02:53
Bunuel What does cannons will hit the target after one round of fire? mean? Irrelevant to question...Was bit tricked!!
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Joined: 02 Sep 2009
Posts: 47983
Re: M01-04  [#permalink]

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27 Jan 2018, 03:08
NandishSS wrote:
Bunuel What does cannons will hit the target after one round of fire? mean? Irrelevant to question...Was bit tricked!!

The question asks to find the probability that none of the three cannons will hit the target after each of them fire once (one round of fire).

Hope it's clear.
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Re: M01-04 &nbs [#permalink] 27 Jan 2018, 03:08
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# M01-04

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