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The police inspector fires 10 shots at the terrorist. If the pr

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The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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New post 19 Sep 2018, 06:50
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A
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C
D
E

Difficulty:

  25% (medium)

Question Stats:

69% (00:46) correct 31% (00:48) wrong based on 29 sessions

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The police inspector fires \(10\) shots at the terrorist. If the probability that he hits the terrorist on any attempt is \(\frac{1}{7}\), find the probability that the terrorist is hit.

A) \(\frac{1}{7}* (\frac{6}{7})^9\)

B) \(\frac{6}{7}* (\frac{1}{7})^9\)

C) \(1- (\frac{1}{7})^{10}\)

D) \(1- (\frac{6}{7})^{10}\)

E) \((\frac{1}{7})^{10}\)
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Re: The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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New post 19 Sep 2018, 07:47
Bismarck wrote:
The police inspector fires \(10\) shots at the terrorist. If the probability that he hits the terrorist on any attempt is \(\frac{1}{7}\), find the probability that the terrorist is hit.

A) \(\frac{1}{7}* (\frac{6}{7})^9\)

B) \(\frac{6}{7}* (\frac{1}{7})^9\)

C) \(1- (\frac{1}{7})^{10}\)

D) \(1- (\frac{6}{7})^{10}\)

E) \((\frac{1}{7})^{10}\)


Probability of hitting in an attempt = 1/7.

So probability of missing in an attempt = 1- (1/7) = 6/7.

Probability of hitting in one of the 10 attempts = 1 - (probability of missing in all attempts) = 1- (6/7)^10.

Hence D.

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Re: The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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New post 19 Sep 2018, 08:13
Clear (D) as explained by Diwakar003
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Re: The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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New post 19 Sep 2018, 08:15
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Bismarck wrote:
The police inspector fires \(10\) shots at the terrorist. If the probability that he hits the terrorist on any attempt is \(\frac{1}{7}\), find the probability that the terrorist is hit.

A) \(\frac{1}{7}* (\frac{6}{7})^9\)

B) \(\frac{6}{7}* (\frac{1}{7})^9\)

C) \(1- (\frac{1}{7})^{10}\)

D) \(1- (\frac{6}{7})^{10}\)

E) \((\frac{1}{7})^{10}\)


ASIDE: If P(shot HITS terrorist) = 1/7, then P(shot MISSES terrorist) = 1 - 1/7 = 6/7

P(terrorist is hit) = P(at least one shot hits)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(having at least 1 shot hit) = 1 - P(not having at least 1 shot hit)
What does it mean to not have at least 1 shot hit? It means having zero shots hit.

So, we can write: P(having at least 1 shot hit) = 1 - P(having zero shots hit)

P(having zero shots hit)
P(having zero shots hit) = P(1st shot misses AND 2nd shot misses AND 3rd shot misses AND . . . AND 10th shot misses
= P(1st shot misses) x P(2nd shot misses) x P(3rd shot misses) x . . . x P(10th shot misses)
= 6/7 x 6/7 x 6/7 x . . . x 6/7
= (6/7)^10

So, P(having at least 1 shot hit) = 1 - P(not having at least 1 shot hit)
= 1 - (6/7)^10

Answer: D

Cheers,
Brent
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Re: The police inspector fires 10 shots at the terrorist. If the pr &nbs [#permalink] 19 Sep 2018, 08:15
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