An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at the $${1^{st}},\,{2^{nd}},\,{3^{rd}}$$ and $${4^{th}}$$ shots are $$0.4,\,0.3,\,0.2$$ and $$0.1$$ respectively. What is the probability that at least one shot hits the plane ?
A.
$$0.6976$$
B.
$$0.3024$$
C.
$$0.72$$
D.
$$0.6431$$
Answer :
$$0.6976$$
Solution :
Probability that at least one shot hits the plane
$$\eqalign{
& = 1 - P\left( {{\text{none of the shot hits the plane}}} \right) \cr
& = 1 - 0.6 \times 0.7 \times 0.8 \times 0.9 \cr
& = 1 - 0.3024 \cr
& = 0.6976 \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$