Question

A shooter can hit a given target with probability \[\dfrac{1}{4}\] . She keeps firing a bullet at the target until she hits it successfully three times and then stops firing. The probability that she fires exactly six bullets lies in the interval.
(a) (0.5272, 0.5274)
(b) (0.2636, 0.2638)
(c) (0.1317, 0.1319)
(d) (0.0658, 0.0660)

Answer 1

Hint: In this question, we first need to know what is meant by binomial distribution. Then check the possibility for the last bullet to be hit correctly which means that the sixth bullet will be the third bullet which will hit the target. Hence, we need to find the probability of hitting twice in the first five bullets.
\[P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\]

Complete step by step answer:
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]
BERNOULLI TRIAL: In a random experiment, if there are any two events, "Success and Failure" and the sum of the probabilities of these two events is 1, then any outcome of such experiment is known as BERNOULLI Trial.
The probability of r successes in n independent BERNOULLI trials is denoted by
\[P\left( X=r \right)\] and is given by
\[P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\]
Where, p = probability of successes
              q = probability of failure and p + q =1.
Let us assume that the probability of hitting the target as p and probability of not hitting the target as q.
As we know that she hits the target at the 6th bullet. So, the possibility is she needs to hit the target twice using the 5 bullets.
\[n=5,r=2\]
\[\Rightarrow p=\dfrac{1}{4}\]
\[\begin{align}
  & \Rightarrow p+q=1 \\
 & \Rightarrow q=1-p \\
 & \Rightarrow q=1-\dfrac{1}{4} \\
 & \therefore q=\dfrac{3}{4} \\
\end{align}\]
Now, the probability will be:
\[\Rightarrow P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\]
On substituting the respective values in the above equation we get,
 \[\begin{align}
  & \Rightarrow P\left( X=2 \right)={}^{5}{{C}_{2}}{{\left( \dfrac{1}{4} \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{5-2}} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{5!}{3!2!}{{\left( \dfrac{1}{4} \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{3}} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{5\times 4\times {{3}^{3}}}{2\times {{4}^{5}}} \\
 & \therefore P\left( X=2 \right)=0.065917 \\
\end{align}\]

Hence, the correct option is (d).

Note: If p = q , then the probability of r successes in n trials is given by:
\[{}^{n}{{C}_{r}}{{p}^{n}}\]
If the total number of trials is n in any attempt and if there are N such attempts, then the total number of r successes is :
\[N\left( {}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}} \right)\]
Here, there is a possibility for taking n =6, r = 3. But, already it is given in the question that she fires until she hits the target successfully 3 times which means that the sixth bullet will be the third bullet which will hit the target.