made easy

Question

The probability of a man hitting a target in one fire is  1/4  The number of times at least he must fire at the target in order that his chance of hitting the target at least once will exceed  2/3  will be _______.

Comments

4 times?.

---

how please explain

---

This can be done other way round. Calculate probability of not hitting the target at all.

Suppose you throw n times .Probability of not hitting the target will be  $_{0}^{n}\textrm{C}(1/4)^{0}(3/4)^{n}$.

Now hitting the target will be 1- $_{0}^{n}\textrm{C}(1/4)^{0}3(/4)^{n}$. So make 1-$_{0}^{n}\textrm{C}1/4^{0}3/4^{n}$>=2/3.

You will get 4 as minimum value satisfying the equation.

---

my dout is this

after solving

(3/4)^n<=1/3

if taking log both side then

n log3/4<=log1/3

n<=log 0.33/0.75

n<=3.85

so ans is 3

is this wrong procedure of solving  this kind of equation

i know if i put n =4 then

(3/4)^4=0.31 which is less then 0.333

so ans must be 4

but i want know  my approach of  solving this kind of equation is correct or not

---

You are doing the opposite.Your n=3 means that his probability of hitting cannot exceed 2/3 until he fires 4 times.

In 1,2,3 attempts he can't have hitting probability greater than 2/3.

You should be careful using this method as it can get you confused. Try with more simple way.

---

ok