Question

A marksman is equally likely to hit or miss his target. How many times must he fire to have a 90% chance to hit the target at least once?  [4 MARKS]

Answer 1

equally likely to hit or miss means probability of hitting is equal to probability of missing . This is possible only when P(Miss) = P(hit) = 1/2 [ because total number of events is only two hit and miss ]

Now, Let he fires n times .
Then probability of missing all times = [P(miss) ]ⁿ = (1/2)ⁿ
Now, question said,
P( at least one hit) = 90% = 90/100 = 0.9

We know, one things ,
P(E) + P(E') = 1 , use it here,
Then, p(at least one hit ) + P(miss) = 1
⇒0.9 + (1/2)ⁿ = 1
⇒ (1/2)ⁿ = 0.1
Take log both sides,
⇒ nlog2 = 1
Log2 = 0.301 so, n = 3.3
So, n = 4

Hence, he fires 4 times to have a 90% chance to hit the target at least once .