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., $$P(I)$$ = 0.3 probability of scoring a target correctly by aeroplane $$II,$$ i.e. $$P(II)$$ = 0.2
$$\therefore \,\,P\left( {\overline I } \right) = 1 - 0.3 = 0.7$$
∴ The required probability
$$\eqalign{
& = P\left( {\overline I \cap II} \right) = P\left( {\overline I } \right).P\left( {II} \right) \cr
& = 0.7 \times 0.2 \cr
& = 0.14 \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$