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M01-04

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M01-04  [#permalink]

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New post 16 Sep 2014, 00:14
1
9
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

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

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Re M01-04  [#permalink]

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New post 16 Sep 2014, 00:14
1
1
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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Re: M01-04  [#permalink]

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New post 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!
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Re: M01-04  [#permalink]

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New post 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!
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Re: M01-04  [#permalink]

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New post 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.
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Re: M01-04  [#permalink]

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New post 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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Re: M01-04  [#permalink]

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New post 27 Jan 2018, 03:08
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Re: M01-04  [#permalink]

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