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Probability Part 1

Section: Mathematics

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Question : 17 of 100

Marks: +1, -0

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is

[26 Feb 2021 Shift 1]

$\; \frac{15}{2^{13}}$
$\; \frac{15}{2^{12}}$
$\; \frac{15}{2^{8}}$
$\; \frac{15}{2^{14}}$

Solution:

Let the coin is tossed '

n

' times.

Also, let

x

denote number of times head occurs.

According to the question,

P(X=7)=P(X=9)

Using formula

P(X=r)={}^{n}C_{t} p^{t} q^{n-r}

where,

p=

probability of getting head in tossing

a

coin

=\frac{1}{2}

and

q=1-p=1-\frac{1}{2}=\frac{1}{2}

\therefore P(X=7)=P(X=9)

\Rightarrow \;\; {}^{n}C_{7} \left(\frac{1}{2}\right)^{7} \left(\frac{1}{2}\right)^{n-7} = {}^{n}C_{9} \left(\frac{1}{2}\right)^{9} \left(\frac{1}{2}\right)^{n-9}

\Rightarrow \;\; {}^{n}C_{7} \left(\frac{1}{2}\right)^{n} = {}^{n}C_{9} \left(\frac{1}{2}\right)^{n}

\Rightarrow \;\; {}^{n}C_{7} = {}^{n}C_{9}

\Rightarrow n=7+9=16[ \because {}^{n}C_{r}={}^{n}C_{q}

only when

n=r+q]

\therefore \;\; P(X=2)={}^{16}C_{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{14}

P(X=2)= \; \frac{16!}{2! 14!} \left(\frac{1}{2}\right)^{16} = \; \frac{16 \times 15}{2} \left(\frac{1}{2}\right)^{16}

= \; \frac{2^{3} \times 15}{2^{16}} = \; \frac{15}{2^{13}}

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