A certain type of missile hits the target with probability $$p = 0.3.$$ What is the least number of missiles should be fired so that there is at least an $$80\% $$ probability that the target is hit ?
A.
$$5$$
B.
$$6$$
C.
$$7$$
D.
none of the above
Answer :
$$5$$
Solution :
Probability of hitting the target $$ = 0.3$$
If $$'n'$$ is the number of times that the Missile is fired.
$$\therefore $$ Probability of hitting at least once
$$\eqalign{
& \Rightarrow 1 - {\left[ {1 - 0.3} \right]^n} = 0.8 \cr
& \Rightarrow {0.7^n} = 0.2 \cr
& \Rightarrow n\,\log \,0.7 = \log \,0.2 \cr
& \Rightarrow n = 4.512 \cr
& {\text{for }}n = 4\,;\,p < 0.8 \cr
& {\text{taken }}n = 5 \cr
& \boxed{n = 5} \cr} $$
Hence $$5$$ missiles should be fired so that there is at least $$80\% $$ probability that the target is hit.
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$