Question

N coins, each with probability $p$ of getting head are tossed together. In the options q = 1-p, The probability of getting odd number of heads is

• A. $\frac{1-(q-p{)}^n}{2}$
• B. $\frac{1+(q-p{)}^n}{2}$
• C. $\frac{(q-p{)}^n}{2}$
• D. $\frac{(q+p{)}^n}{2}$

Answer 1

Answer: (A)

$\frac{1-(q-p{)}^n}{2}$

In the expansion of $(x+a{)}^n$

if the sum of odd - term be 'P' and sum of even be'Q'

$(x+a{)}^n=x^n{+}^n{C}_1{x}^{n-1}a{+}^n{C}_2{x}^{n-2}{a}^2{+}^n{C}_3{x}^{n-3}{a}^3+...........{+}^n{C}_n{a}^n$

$=(n^n{+}^n{C}_2{x}^{n-2}{a}^2+........)+{(}_1^C{x}^{n-1}a{+}^n{C}_3{x}^{n-3}{a}^3+.........)$

$(x+a{)}^n=P+Q------->(1)$ and

$(x-a{)}^n$ = $x^n{-}^n{C}_1{x}^{n-1}a{+}^n{C}_2{x}^{n-2}{a}^2{-}^n{C}_3{x}^{n-3}{a}^3+......+(-1{)}^{nn}{C}_n{a}^n$

$=(x^n{+}^n{C}_2{x}^n-2a^2+......)-{(}^n{C}_1{x}^{n-1}a{+}^n{C}_3{x}^{n-3}{a}^3+........)$

$(x-a{)}^n$= $P-Q$----------> (2)

Hence sum of t add terms =P=$(x+a{)}^n-(x-a{)}^n$

Probability of getting head = p
Probability of getting tail = 1-p =q
Now, P(odd no. of heads in N tosses = P(1 head)+P(3 heads) +P(5 heads)+.....
${=}^N{C}_{1}p{q}^{N-1}{+}^N{C}_3{p}^3{q}^{N-3}{+}^N{C}_5{p}^5{q}^{N-5}+..$

$=\frac{(p+q{)}^N-(p-q{)}^N}{2}$
Using the property of binomial theorem
as shown in the image below
since $p+q=1$
$=\frac{1-(p-q{)}^N}{2}$