Question

A missile target may be at a point P with probability $\dfrac{9}{{10}}$ or at point Q with probability $\dfrac{1}{{10}}$ we have 20 shells each of which can be fired either at point P or Q. Each shell may hit the target independently of the other shoot with probability $\dfrac{2}{3}$. Then the number of shells must be fired at point P to hit any target with maximum probability is
A. 10
B. 11
C. 12
D. 13

Answer 1

This question talks about the probability of two different events or species which stance to occur differently. The extent to which each event occurred is expressed as ratios and proportions as the exact numbers of the items will be solved for.

Complete step by step answer:
We are given that the probability of the missile at P is $\dfrac{9}{{10}}$, and at Q is $\dfrac{1}{{10}}$ and there are 20 shells. Each shell may hit the target independently of the other shoot with probability $\dfrac{2}{3}$.
Let A be the event that the target gets hit, E be the event to select target P and F be the event to select target Q.
$P\left( E \right) = \dfrac{9}{{10}} \\\ P\left( F \right) = \dfrac{1}{{10}} \\\ P\left( {E \cup F} \right) = \dfrac{2}{3} \\\$
Using Conditional probability, Probability that target gets hit only when the target is P and probability that target gets hit only when the target is Q are given below respectively.
$P\left( {\dfrac{A}{E}} \right) = 1 - {\left( {\dfrac{1}{3}} \right)^x} \\\ P\left( {\dfrac{A}{F}} \right) = 1 - {\left( {\dfrac{1}{3}} \right)^{20 - x}} \\\$
Where x is the no. of shells used to point at target P
Therefore,
$P\left( A \right) = P\left( E \right).P\left( {\dfrac{A}{E}} \right) + P\left( F \right).P\left( {\dfrac{A}{F}} \right) \\\ \Rightarrow P\left( A \right) = \dfrac{9}{{10}} \times \left( {1 - {{\left( {\dfrac{1}{3}} \right)}^x}} \right) + \dfrac{1}{{10}} \times \left( {1 - {{\left( {\dfrac{1}{3}} \right)}^{20 - x}}} \right) \\\$
P(A) is a probability which is a constant. Differentiating with respect to x, we get
$\Rightarrow 0 = \left[ {\dfrac{9}{{10}} \times \left( { - {{\left( {\dfrac{1}{3}} \right)}^x}} \right) \times - {{\log }_e}3} \right] + \left[ {\dfrac{1}{{10}} \times {{\left( {\dfrac{1}{3}} \right)}^{20 - x}} \times - {{\log }_e}3} \right] \\\ \Rightarrow \dfrac{{{{\log }_e}3}}{{10}}\left[ {\left( {9 \times {3^{ - x}}} \right) - {3^{x - 20}}} \right] = 0 \\\ \Rightarrow \left( {{3^2} \times {3^{ - x}}} \right) - {3^{x - 20}} = 0 \\\ \Rightarrow {3^{2 - x}} = {3^{x - 20}} \\\ \Rightarrow 2 - x = x - 20 \\\ \Rightarrow x + x = 20 + 2 \\\ \Rightarrow 2x = 22 \\\ \therefore x = 11 \\\$
Therefore, the no. of shells that must be fired at point P to hit any target with maximum probability is 11.
Therefore, Option B is correct.

Note: Probability (may not hit the target at P) = $1 - \dfrac{9}{{10}} = \dfrac{{10 - 9}}{{10}} = \dfrac{1}{{10}}$ and the Probability (may not hit the target at Q) = $1 - \dfrac{1}{{10}} = \dfrac{{10 - 1}}{{10}} = \dfrac{9}{{10}}$. For every event to occur or number of events to occur we can always get them from the universal set. Probability is not a possibility. Possibility is a qualitative characteristic of an event whereas probability of an event is the likelihood with which that event could happen.