A coin is tossed $$7$$ times. Each time a man calls head. The probability that he wins the toss on more occasions is :
A.
$$\frac{1}{4}$$
B.
$$\frac{5}{8}$$
C.
$$\frac{1}{2}$$
D.
none of these
Answer :
$$\frac{1}{2}$$
Solution :
The man has to win at least $$4$$ times.
$$\therefore $$ the required probability
$$\eqalign{
& = {}^7{C_4}.{\left( {\frac{1}{2}} \right)^4}.{\left( {\frac{1}{2}} \right)^3} + {}^7{C_5}.{\left( {\frac{1}{2}} \right)^5}.{\left( {\frac{1}{2}} \right)^2} + {}^7{C_6}.{\left( {\frac{1}{2}} \right)^6}.\frac{1}{2} + {}^7{C_7}{\left( {\frac{1}{2}} \right)^7} \cr
& = \left( {{}^7{C_4} + {}^7{C_5} + {}^7{C_6} + {}^7{C_7}} \right).\frac{1}{{{2^7}}} \cr
& = \frac{{64}}{{{2^7}}} \cr
& = \frac{1}{2} \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$