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

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Senior Manager
Joined: 18 Jun 2018
Posts: 268
The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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19 Sep 2018, 06:50
5
00:00

Difficulty:

35% (medium)

Question Stats:

64% (00:58) correct 36% (01:15) wrong based on 50 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) $$(\frac{1}{7})^{10}$$
Manager
Joined: 02 Aug 2015
Posts: 153
Re: The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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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.

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

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19 Sep 2018, 08:13
Clear (D) as explained by Diwakar003
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Abhishek....

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

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19 Sep 2018, 08:15
Top Contributor
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

Cheers,
Brent
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Joined: 26 Mar 2018
Posts: 14
Re: The police inspector fires 10 shots at the terrorist. If the pr  [#permalink]

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02 Jul 2019, 11:33
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) $$(\frac{1}{7})^{10}$$

Re: The police inspector fires 10 shots at the terrorist. If the pr   [#permalink] 02 Jul 2019, 11:33
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