First slide

Random variable and probability distribution

Question

A fair coin is tossed 100 times. The probability of getting tails 1, 3, ……., 49 times is

Moderate

A

1/2
B

1/4
C

1/8
D

1/16

Solution

Let the probability of getting a tail in a single trial be p = 1/2. The number of trials be n = 100 and the number of trials in 100 trials be X.
We have,
$P (X = r) =^{100} C_{n} p^{r} q^{n - r}$
$=^{100} C_{r} {(\frac{1}{2})}^{r} {(\frac{1}{2})}^{100 - r} =^{100} C_{r} {(\frac{1}{2})}^{100}$
Now,
$P (X = 1) + P (X = 3) + \dots + P (X = 49)$
$\begin{array}{l} =^{100} C_{1} {(\frac{1}{2})}^{100} +^{100} C_{3} {(\frac{1}{2})}^{100} + \dots +^{100} C_{49} {(\frac{1}{2})}^{100} \\ = (^{100} C_{1} +^{100} C_{3} + \dots +^{100} C_{49}) {(\frac{1}{2})}^{100} \end{array}$
But $^{100} C_{1} +^{100} C_{3} + \dots +^{100} C_{99} = 2^{99}$
Also, $^{100} C_{99} =^{100} C_{1}$
$^{100} C_{97} =^{100} C_{3}, \dots^{100} C_{51} =^{100} C_{49}$
Thus, $2 (^{100} C_{1} +^{100} C_{3} + \dots +^{100} C_{49}) = 2^{99}$
or $^{100} C_{1} +^{100} C_{3} + \dots +^{100} C_{49} = 2^{98}$
Therefore, probability of required event is
$\frac{2^{98}}{2^{100}} = \frac{1}{4} 2^{98} / 2^{100} = 1 / 4$

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