Question

hey ☺
.....…....
⭐⭐⭐⭐⭐⭐⭐☺☺☺⭐⭐⭐⭐⭐⭐⭐⭐


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

so the answer is 4 he have to fire 4 times this is Ur required result

Answer 2

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 .

2nd method :-
Use formula $\bold{P(x=r)=^nC_rP(E)^rP(E')^{n-r}}}$
Here P(E) denotes probability of hitting
P(E') denotes probability of missing
r denotes how many times of hitting

Now, for at least once hitting , r = 1
P(x = 1) = 0.9 [ as given 90% ]
but we know, P(x= 1) + P(x = 0) = 1
So, P(x = 1) = 1 - P(x = 0)
0.9 = 1 - ⁿC₀(1/2)⁰(1/2)ⁿ⁻⁰ = 1 - 1/2ⁿ
Now, solve then n = 3.3
So, 4 times he fires