Question

The probability that a person will hit a target in shooting is 0.3. If he shoots 10 times, then probability of his shooting the target is
A) $1$
B) $1 – {0.7}^{10}$
C) ${0.7}^{10}$
D) ${0.3}^{10}$

Answer 1

Using the axiomatic approach of probability we shall create an expression. This expression shall be the intersection of 10 events which correspond to hitting of the target by the person. Then we shall find the probability of intersection of events.

Complete step by step solution:
Let $E$ be the event of hitting the target by the person.
We already know that, according to the axiomatic approach to probability, the probability of an event $E$ is : $0 \leqslant P(E) \leqslant 1$
So according to question we have $P(E) = 0.3$
Now, the question says the man shoots 10 times. So, if he shoots each time then the event $E$ occurs 10 times. This can be represented as
$E$ and $E$ and $E$ and $E$ and $E$ and $E$ and $E$ and $E$ and $E$ and $E$
Now, the fundamental principle of counting states that whenever two events occur together, that is $A$ “and” $B$ occur together the total outcomes are $A \cdot B$
So, the intersection of events we get,
$E \cap E \cap E \cap E \cap E \cap E \cap E \cap E \cap E \cap E \\\ \Rightarrow E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \\\$
We can calculate the probability easily. The Required probability is
$P(E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E \cdot E) \\\ \Rightarrow P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(E) \\\ \Rightarrow P{(E)^{10}} \\\ \Rightarrow {(0.3)^{10}} \\\$

So, the probability of hitting the target ten times is (0.3)10
Hence, the correct option is D.

Note:
While applying the fundamental principle of counting always make sure of the condition stated. We use addition in case of ‘OR’ condition and multiplication in case of ‘AND’ condition.