Two aeroplanes $$I$$ and $$II$$ bomb a target in succession. The probabilities of $$I$$ and $$II$$ scoring a hit correctly are $$0.3$$ and $$0.2,$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is :
A.
$$0.2$$
B.
$$0.7$$
C.
$$0.06$$
D.
$$0.14$$
Answer :
$$0.14$$
Solution :
Given : Probability of aeroplane $$I$$, scoring a target correctly i.e., $$P\left( I \right) = 0.3$$ probability of scoring a target correctly by aeroplane $$II$$, i.e. $$P\left( {II} \right) = 0.2$$
$$\eqalign{
& \therefore \,P\left( {\overline I } \right) = 1 - 0.3 = 0.7 \cr
& \therefore \,{\text{The required probability}} \cr
& = P\left( {\overline I \cap II} \right) \cr
& = P\left( {\overline I } \right).P\left( {II} \right) \cr
& = 0.7 \times 0.2 \cr
& = 0.14 \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$