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Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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9 00:00

Difficulty:   5% (low)

Question Stats: 78% (00:56) correct 22% (01:12) wrong based on 183 sessions

### HideShow timer Statistics 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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Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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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.

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Intern  Joined: 04 Sep 2015
Posts: 35
Location: Germany
Concentration: Operations, Finance
WE: Project Management (Aerospace and Defense)

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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.

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: 110
Location: United States
GMAT 1: 720 Q50 V38 GPA: 3.8
WE: Information Technology (Computer Software)

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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.

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: 35
Location: Germany
Concentration: Operations, Finance
WE: Project Management (Aerospace and Defense)

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[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.
Director  V
Joined: 06 Jan 2015
Posts: 680
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)

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Bunuel What does cannons will hit the target after one round of fire? mean? Irrelevant to question...Was bit tricked!!
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Resource: GMATPrep RCs With Solution
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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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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Manager  B
Joined: 27 Nov 2015
Posts: 123

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why cant we add the probabilities on not happening Re: M01-04   [#permalink] 20 Feb 2019, 08:10
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# M01-04

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