Question

A fair coin is tossed 100 times. The sum of the probabilities of getting tails 1, 3, 5,…., 49 times is
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{8}$
D. $\dfrac{1}{16}$

Answer 1

Hint: In this question, as the coin is fair, we can derive that the probability of getting a tail in one toss is$\dfrac{1}{2}$ and we use binomial distribution to get the answer.
P (k tails in 100 tosses) =${}^{100}{{C}_{k}}{{\left( \dfrac{1}{2} \right)}^{k}}{{\left( \dfrac{1}{2} \right)}^{100-k}}$=${}^{100}{{C}_{k}}{{\left( \dfrac{1}{2} \right)}^{100}}=\dfrac{{}^{100}{{C}_{k}}}{{{2}^{100}}}$
The required probability is the summation of the probabilities of the tails occurring 1,3,5…49 tosses, which is
$\dfrac{{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}...........{}^{100}{{C}_{49}}}{{{2}^{100}}}$which we can obtain using the formulae stated below.
\[{}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}.....+{}^{n}{{C}_{n\text{ or n-1}}}={{2}^{n-1}}\] $\text{where }{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
After using this, we can find the required probability using the relation\[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]

Complete step-by-step answer:
In the question, it is given that a fair coin is tossed 100 times. We can conclude that the probability of getting a head is the same as that of tail as the coin is fair.
$\therefore $P (Tail in a toss) = P (Head in a toss) and the sum of the two probabilities is 1.
$\therefore $ P (Tail in a toss) = P (Head in a toss) = $\dfrac{1}{2}$
Let us consider an event E for which the success is getting ‘k’ tails out of 100 tosses. We have to find the probability of success of the event E. We have to select ‘k’ tosses out of 100 tosses and multiply with the probabilities of success and failure of getting a tail to get the required probability.
P (Success of E) = P (k tails in 100 tosses) = ${}^{100}{{C}_{k}}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.......\text{k times }\text{.}\dfrac{1}{2}.\dfrac{1}{2}.........100\text{-k times}$
$\Rightarrow $ P (k tails in 100 tosses) =${}^{100}{{C}_{k}}{{\left( \dfrac{1}{2} \right)}^{k}}{{\left( \dfrac{1}{2} \right)}^{100-k}}$=${}^{100}{{C}_{k}}{{\left( \dfrac{1}{2} \right)}^{100}}=\dfrac{{}^{100}{{C}_{k}}}{{{2}^{100}}}\to (1)$
We use the equation-1 to solve the problem.
P (1 tail in 100 tosses) = $\dfrac{{}^{100}{{C}_{1}}}{{{2}^{100}}}$
P (3 tails in 100 tosses) = $\dfrac{{}^{100}{{C}_{3}}}{{{2}^{100}}}$
Likewise, P (49 tails in 100 tosses)= $\dfrac{{}^{100}{{C}_{49}}}{{{2}^{100}}}$
We have to add the probabilities to get the required probability. Adding them, we get
Required probability=$\dfrac{{}^{100}{{C}_{1}}}{{{2}^{100}}}+\dfrac{{}^{100}{{C}_{3}}}{{{2}^{100}}}+\dfrac{{}^{100}{{C}_{5}}}{{{2}^{100}}}..........\dfrac{{}^{100}{{C}_{49}}}{{{2}^{100}}}=\dfrac{{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}...........{}^{100}{{C}_{49}}}{{{2}^{100}}}\to (2)$
We have to use a formula from the binomial expansions to get the value of the numerator. It is
Formula: \[{}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}.....+{}^{n}{{C}_{n\text{ or n-1}}}={{2}^{n-1}}\]
Substituting n=100 in the above formula, we get
\[{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}.....+{}^{100}{{C}_{n\text{ or n-1}}}={{2}^{99}}\to (3)\]
But we require the sum up to${}^{100}{{C}_{49}}$. To get this, we use the property of the combinations which is\[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\] which means that
${}^{100}{{C}_{1}}={}^{100}{{C}_{99}}\text{ ; }{}^{100}{{C}_{3}}={}^{100}{{C}_{97}}\text{ ; }{}^{100}{{C}_{47}}={}^{100}{{C}_{53}}\text{ ; }{}^{100}{{C}_{49}}={}^{100}{{C}_{51}}$
Using the above property in equation-3 we get
\[{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}+.....{}^{100}{{C}_{47}}+{}^{100}{{C}_{49}}+{}^{100}{{C}_{49}}+{}^{100}{{C}_{47}}.........{}^{100}{{C}_{3}}+{}^{100}{{C}_{1}}={{2}^{99}}\to (4)\]
We can see that every term in the summation has repeated twice. So,
\[2\times \left( {}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}+.....{}^{100}{{C}_{47}}+{}^{100}{{C}_{49}} \right)={{2}^{99}}\]
Cancelling the 2 gives
\[{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}+.....{}^{100}{{C}_{47}}+{}^{100}{{C}_{49}}={{2}^{98}}\]
Using the obtained result in equation-2, we get
Required probability $=\dfrac{{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}+.....{}^{100}{{C}_{47}}+{}^{100}{{C}_{49}}}{{{2}^{100}}}=\dfrac{{{2}^{98}}}{{{2}^{100}}}$
Cancelling the numerator with denominator, we get
Required probability$=\dfrac{1}{4}$
$\therefore $The answer is option B.

Note:Students, who know the formula used from the binomial expansion partially, can commit a mistake by considering \[{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+{}^{100}{{C}_{5}}+.....{}^{100}{{C}_{47}}+{}^{100}{{C}_{49}}={{2}^{99}}\] which is wrong. Instead, anyone can derive the formula using the relation as follows${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{\left( 1 \right)}^{n}}{{\left( x \right)}^{0}}+{}^{n}{{C}_{1}}{{\left( 1 \right)}^{n-1}}{{\left( x \right)}^{1}}+{}^{n}{{C}_{2}}{{\left( 1 \right)}^{n-2}}{{\left( x \right)}^{2}}.......{}^{n}{{C}_{n}}{{\left( 1 \right)}^{0}}{{\left( x \right)}^{n}}$. Substituting x=1 and x= -1 in the above expansion, we get
\[\begin{align}
  & {{2}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+......{}^{n}{{C}_{n}} \\
 & 0={}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}-{}^{n}{{C}_{3}}......\pm {}^{n}{{C}_{n}} \\
\end{align}\]
By adding or subtracting, we can get the required results.