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Question: A fair coin is tossed \( 100 \) times. The probability of getting tails on odd number of times is ...

A fair coin is tossed 100100 times. The probability of getting tails on odd number of times is
A. 12\dfrac{1}{2}
B. 14\dfrac{1}{4}
C. 18\dfrac{1}{8}
D. 38\dfrac{3}{8}

Explanation

Solution

Hint : Probability is the state of being probable and the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes. The binomial expansion or the binomial theorem describes the algebraic expansion of the powers of the binomial (binomial is the pair of two terms). Use formula (1+x)n=nC1x1+nC2x2+.....{(1 + x)^n} = {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + ..... for binomial expansion

** Complete step-by-step answer** :
Using - the probability distribution for events occurring n times and the required “q” times.
The probability of getting tail is P(T)=12P(T) = \dfrac{1}{2}
Similarly, the probability of getting tail is P(H)=12P(H) = \dfrac{1}{2}
And tail is required an odd number of times.
Therefore, q=1,3,5,7,9,......99q = 1,3,5,7,9,......99
Now, the required probability is
=P(q=1)+P(q=3)+......+P(q=99)= P(q = 1) + P(q = 3) + ...... + P(q = 99)
Substituting the values in the above equation -
=100C1(12)99(12)1+100C3(12)97(12)3+.....+100C99(12)1(12)99= {}^{100}{C_1}{\left( {\dfrac{1}{2}} \right)^{99}}{\left( {\dfrac{1}{2}} \right)^1} + {}^{100}{C_3}{\left( {\dfrac{1}{2}} \right)^{97}}{\left( {\dfrac{1}{2}} \right)^3} + ..... + {}^{100}{C_{99}}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{99}}
We can observe that power to the (12)\left( {\dfrac{1}{2}} \right) sums up to 100100 . Therefore, take multiple common from all the terms in the bracket.
=(12)100[100C1+100C3+100C5+....+100C99]= {\left( {\dfrac{1}{2}} \right)^{100}}\left[ {{}^{100}{C_1} + {}^{100}{C_3} + {}^{100}{C_5} + .... + {}^{100}{C_{99}}} \right] .... (A)
Now, taking the expansion of the binomial expansion-
(1+x)n=nC0x0+nC1x1+nC2x2+.....+nCnxn{(1 + x)^n} = {}^n{C_0}{x^0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + ..... + {}^n{C_n}{x^n}
Expand the above equation-
(1+x)n=1+nC1x1+nC2x2+.....+nCnxn{(1 + x)^n} = 1 + {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + ..... + {}^n{C_n}{x^n} .... (B)
Place x=1x = 1 in the above equation –
(1+1)n=1+nC1+nC2+.....+nCn{(1 + 1)^n} = 1 + {}^n{C_1} + {}^n{C_2} + ..... + {}^n{C_n}
⇒(2)n=1+nC1+nC2+.....+nCn\Rightarrow {(2)^n} = 1 + {}^n{C_1} + {}^n{C_2} + ..... + {}^n{C_n} ..... (C)
Place x=(−1)x = ( - 1) in the equation (B)
(1−1)n=1−nC1+nC2−.....+nCn{(1 - 1)^n} = 1 - {}^n{C_1} + {}^n{C_2} - ..... + {}^n{C_n}
⇒0=1−nC1+nC2−.....+nCn\Rightarrow 0 = 1 - {}^n{C_1} + {}^n{C_2} - ..... + {}^n{C_n} .... (D)
Here, n=100n = 100 . Subtract equation (D) from (C)
⇒(2)100=2100C1+2100C3+.....+2100C99\Rightarrow {(2)^{100}} = 2{}^{100}{C_1} + 2{}^{100}{C_3} + ..... + 2{}^{100}{C_{99}}
Take multiple common from all the terms on the right hand side of the equation.
⇒(2)100=2[100C1+100C3+.....+100C99]\Rightarrow {(2)^{100}} = 2[{}^{100}{C_1} + {}^{100}{C_3} + ..... + {}^{100}{C_{99}}]
⇒(2)1002=[100C1+100C3+.....+100C99]\Rightarrow \dfrac{{{{(2)}^{100}}}}{2} = [{}^{100}{C_1} + {}^{100}{C_3} + ..... + {}^{100}{C_{99}}] .... (E)
Place the above value in equation (A)
=(12)100×(2)1002= {\left( {\dfrac{1}{2}} \right)^{100}} \times \dfrac{{{{(2)}^{100}}}}{2}
Simplify the above equation –
=(12100)×(2)1002= \left( {\dfrac{1}{{{2^{100}}}}} \right) \times \dfrac{{{{(2)}^{100}}}}{2}
Common multiples from the numerator and the denominator cancel each other.
=12= \dfrac{1}{2}
So, the correct answer is “ =12= \dfrac{1}{2} ”.

Note : Know the correct formula for the expansion of the binomials. Follow the laws of the power and exponent to simplify the expression. The probability of any event always ranges between zero and one. It can never be the negative number or the number greater than one