A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses is :
A.
$$\frac{1}{2}$$
B.
$$\frac{1}{{32}}$$
C.
$$\frac{{31}}{{32}}$$
D.
$$\frac{1}{{16}}$$
Answer :
$$\frac{1}{{16}}$$
Solution :
Let us find the probability of Heads appearing in the toss since $$P\left( H \right) = P\left( T \right) = \frac{1}{2}$$ for an unbiased coin, it will not affect or introduce any error in the result.
Thus
If a coin is tossed Once.
$$P\left( H \right) = \frac{1}{2}$$
If a coin is tossed twice.
The probability of head appearing twice is $$ = \frac{1}{2}.\frac{1}{2} = \frac{1}{4}$$
If a coin is tossed thrice.
The probability of head appearing thrice is $$ = \frac{1}{2}.\frac{1}{2}.\frac{1}{2} = \frac{1}{8}$$
If a coin is tossed 4 times.
The probability of head appearing fourth times is $$ = \frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2} = \frac{1}{{16}}$$
Therefore we can generalize.
If coin is tossed $$n$$ times.
The probability of head appearing $$n$$ times will $$ = \frac{1}{2}.\frac{1}{2}........ = \frac{1}{{{2^n}}}$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$