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 =

6/7P(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)^10So, P(having at least 1 shot hit) = 1 -

P(not having at least 1 shot hit)= 1 -

(6/7)^10Answer: D

Cheers,

Brent

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