Question

A fair coin is tossed 100 times the probability of getting tails an odd number of time is
a) $\dfrac{1}{2}$
b) $\dfrac{1}{4}$
c) 0
d) 1

Answer 1

Here the given question is based on the concept of probability. We have to find the probability of getting tails an odd number of times. For this first we have to find the probability of head and tail. Further by using a binomial distribution i.e., $P\left( {X = x} \right){ = ^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$, to get the required probability.

Complete step by step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the
number of favourable outcomes and the total number of outcomes.
$\text{Probability of event to happen}\,P\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total Number of outcomes}}}$
Consider the given question,
When a coin is tossed we get the output as head or tail. The probability of head is $\dfrac{1}{2}$
Let us take ‘$p$’ be the probability of head and ‘$q$’ be the probability of tail.
Probability of head $p = \dfrac{1}{2}$ ----(1)
Probability of tail $q = 1 - p = 1 - \dfrac{1}{2} = \dfrac{1}{2}$ ----(2)
Let $X$ be the random variable that represents the tail in 100 trials. It’s a Bernoulli trial so, $X$ has a binomial distribution, we obtain
$P\left( {X = x} \right){ = ^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$ ----(3)
Probability of getting tails an odd number $= P\left( {X \geqslant 1} \right)$
So we have
$\Rightarrow P\left( {X = 1} \right) + P(X = 3) + P(X = 5) + ... + P(X = 99)$
On putting values of $p$, $q$, $n$ and $x$ in equation (3), then
Required probability ${ = ^{100}}{C_1}\left( {\dfrac{1}{2}} \right){\left( {\dfrac{1}{2}} \right)^{99}} + {\,^{100}}{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{97}} + ... + {}^{100}{C_{99}}{\left( {\dfrac{1}{2}} \right)^{99}}{\left( {\dfrac{1}{2}} \right)^1}$
On simplifying we have
${ \Rightarrow ^{100}}{C_1}{\left( {\dfrac{1}{2}} \right)^{100}} + {\,^{100}}{C_3}{\left( {\dfrac{1}{2}} \right)^{100}} + ... + {}^{100}{C_{99}}{\left( {\dfrac{1}{2}} \right)^{100}}$
If a number 1 has any power then the result will be 1.
${ \Rightarrow ^{100}}{C_1}\dfrac{1}{{{2^{100}}}} + {\,^{100}}{C_3}\dfrac{1}{{{2^{100}}}} + ... + {}^{100}{C_{99}}\dfrac{1}{{{2^{100}}}}$
$\Rightarrow \dfrac{{^{100}{C_1}{ + ^{100}}{C_3} + ...{ + ^{100}}{C_{99}}}}{{{2^{100}}}}$
We know the formula of combination$^n{C_1} + {}^n{C_3} + {}^n{C_5} + .... = {2^{n - 1}}$
$\Rightarrow \dfrac{{{2^{100 - 1}}}}{{{2^{100}}}}$
We know the property of exponent ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$, then
$\Rightarrow \dfrac{{\dfrac{{{2^{100}}}}{2}}}{{{2^{100}}}}$
By taking reciprocal we have
$\Rightarrow \dfrac{{{2^{100}}}}{2} \times \dfrac{1}{{{2^{100}}}}$
$\Rightarrow \dfrac{1}{2}$
Therefore the option (a) is the correct option.

Note:
The probability is a number of possible values. Candidate must know we have to use
permutation concept or combination concept to solve the given problem because it is the first and main thing to solve the problem.
$1 - P\left( A \right)$ refers to the probability that Event A will not occur is denoted by $P\left( {A'} \right)$.
Remember, factorial is the continued product of first n natural numbers is called the “n factorial “ and it represented by $n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot 3 \cdot 2 \cdot 1$.