probability

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

Answer is 4.

Answer 1

P[Hitting] = $\frac{1}{4}$ ; P[not hitting] = $\frac{3}{4}$

Probability of hitting target at least once in $n$ fire = $P[X \geq 1] = 1- P[X=0]$

It is given:  $P[X\geq1] > \frac{2}{3}$

$\Rightarrow1-P[X=0] > \frac{2}{3}$

$\Rightarrow1-\frac{2}{3}>P[X=0]$

$\Rightarrow\frac{1}{3}>\binom{n}{0}(\frac{1}{4})^{0}(\frac{3}{4})^{n}$

$\Rightarrow\frac{1}{3}>(\frac{3}{4})^{n}$

$\Rightarrow4^{n} > 3^{n+1}$(Don't know how to solve this. Calculated by putting values)

$\Rightarrow n= 4$

Answer 2

P = 1/4  (success)
q = 3/4  (failure)

as probability of success is same trail to trial, only two events success and failure, and infinite number of trials.
We can use binomial distribution
$P(x>=1) = 1 -P(x=0)$
$= 1 - p(x=0) >2/3$
$= 1 -nC0(1/4)^0(3/4)^n > 2/3$
$(3/4)^n < 1/3 $ now check for n=1, 2, 3, 4 ....

answer is 4.