Question

The probability that at least one of the events $$A$$ and $$B$$ occurs is $$\frac{3}{5}.$$ If $$A$$ and $$B$$ occur simultaneously with probability $$\frac{1}{5}$$ then $$P\left( {A'} \right) + P\left( {B'} \right)$$    is :

A. $$\frac{2}{5}$$
B. $$\frac{4}{5}$$
C. $$\frac{6}{5}$$  
D. $$\frac{7}{5}$$
Answer :   $$\frac{6}{5}$$
Solution :
Here, $$P\left( {A \cup B} \right) = \frac{3}{5}$$    and $$P\left( {A \cap B} \right) = \frac{1}{5}.$$    So, from the addition theorem,
$$\eqalign{ & \frac{3}{5} = P\left( A \right) + P\left( B \right) - \frac{1}{5} \cr & {\text{or }}\frac{4}{5} = 1 - P\left( {A'} \right) + 1 - P\left( {B'} \right) \cr & \therefore \,P\left( {A'} \right) + P\left( {B'} \right) = 2 - \frac{4}{5} = \frac{6}{5} \cr} $$

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:

A. Mutually exclusive
B. Independent and mutually exclusive
C. Dependent
D. None of these
Releted Question 2

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

A. 0.39
B. 0.25
C. 0.11
D. none of these
Releted Question 3

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

A. 0.936
B. 0.784
C. 0.904
D. none of these
Releted Question 4

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. $$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$

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