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# The snipers shoot a certain target. Their probabilities of

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CEO
Joined: 21 Jan 2007
Posts: 2734

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Location: New York City
The snipers shoot a certain target. Their probabilities of [#permalink]

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04 Nov 2007, 01:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability:

1. that exactly one sniper missed?
2. that exactly one sniper hit?
3. that exactly two snipers missed?
4. that exactly two snipers hit?

Kudos [?]: 1049 [0], given: 4

Manager
Joined: 01 Nov 2007
Posts: 68

Kudos [?]: 9 [0], given: 0

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04 Nov 2007, 05:42
I feel Q1 and Q4 & Q2 and Q3 are the same. Arent they?

I mean if exactly one missed the other two hit and when exactly one hit the other two missed.

And is the answer of Q1 = 0.485 and that of Q2 = 0.185.

If so I can provide my explanation

Kudos [?]: 9 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1049 [0], given: 4

Location: New York City

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18 Nov 2007, 17:51
Proababilty that Sniper 1 misses and other 2 hit = (1-0.9) (0.7) (0.5) = 0.1 * 0.7 * 0.5

Probability that Sniper 2 misses and other 2 hit = (0.9) (1-0.7) (0.5) = 0.9 * 0.3 * 0.5

Probability that sniper 3 misses and other 2 hit = (0.9) (0.7)(1-0.5) = 0.9 * 0.7 * 0.5

Therefore Probability that exactly 1 sniper misses = .1(.7)(.5)+.9(.7)(.5)+.9(.3)(.5)

see any errors?

Kudos [?]: 1049 [0], given: 4

Director
Joined: 13 Dec 2006
Posts: 506

Kudos [?]: 243 [0], given: 0

Location: Indonesia

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18 Nov 2007, 17:58
yep, I do

Following is my explanation:

1. Exacty one sniper missed = 0.9*0.7*0.5 + 0.9*0.3*0.5 + 0.1*0.7*0.5 = 0.485

2. Exactly one sniper hit= 0.9*0.3*0.5 + 0.1*0.7*0.5 + 0.1*0.3*0.5 =0.185

3. Exactly two sniper missed = Exactly one snipper hit

4. Exactly two snipers hit = one sniper missed

Amar

Kudos [?]: 243 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1049 [0], given: 4

Location: New York City

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03 Dec 2007, 08:22
Amardeep Sharma wrote:
yep, I do

Following is my explanation:

1. Exacty one sniper missed = 0.9*0.7*0.5 + 0.9*0.3*0.5 + 0.1*0.7*0.5 = 0.485

2. Exactly one sniper hit= 0.9*0.3*0.5 + 0.1*0.7*0.5 + 0.1*0.3*0.5 =0.185

3. Exactly two sniper missed = Exactly one snipper hit

4. Exactly two snipers hit = one sniper missed

Amar

very nice.

Kudos [?]: 1049 [0], given: 4

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1049 [0], given: 4

Location: New York City
Re: joint Probability - salvo [#permalink]

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03 Dec 2007, 08:32
The snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability that at least 1 sniper missed?

Kudos [?]: 1049 [0], given: 4

Director
Joined: 25 Oct 2006
Posts: 635

Kudos [?]: 637 [0], given: 6

Re: joint Probability - salvo [#permalink]

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17 Aug 2008, 12:41
(1)that exactly one sniper missed = (4) that exactly two snipers hit = 0.9*0.7*0.5 + 0.9*0.3*0.5 + 0.1*0.7*0.5 = 0.485
(2)that exactly one sniper hit = (3) that exactly two snipers missed = 0.9*0.3*0.5 + 0.1*0.7*0.5 + 0.1*0.3*0.5 =0.185
all three missed = 0.1*0.3*0.5

At least one missed = Exactly one missed + exactly two missed + all three missed = 0.485 + 0.185 + 0.015 = 0.685
_________________

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Kudos [?]: 637 [0], given: 6

Manager
Joined: 14 Jun 2008
Posts: 162

Kudos [?]: 37 [0], given: 0

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18 Aug 2008, 00:01
Amardeep Sharma wrote:
yep, I do

Following is my explanation:

1. Exacty one sniper missed = 0.9*0.7*0.5 + 0.9*0.3*0.5 + 0.1*0.7*0.5 = 0.485

2. Exactly one sniper hit= 0.9*0.3*0.5 + 0.1*0.7*0.5 + 0.1*0.3*0.5 =0.185

3. Exactly two sniper missed = Exactly one snipper hit

4. Exactly two snipers hit = one sniper missed

Amar

should you multply each of the above by 1/3
for e.g.
Exacty one sniper missed = 1/3(0.9*0.7*0.5) + 1/3(0.9*0.3*0.5) + 1/3(0.1*0.7*0.5)
as each of the three events have a probability of occuring 1/3

Kudos [?]: 37 [0], given: 0

Re:   [#permalink] 18 Aug 2008, 00:01
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