Question

Four tickets marked $00,01,10,11$ respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the number on tickets thus drawn is $23,$ is

• A. $\frac{41}{1024}$
• B. $\frac{25}{256}$
• C. $\frac{25}{178}$
• D. $\frac{21}{256}$

Answer 1

The correct option is B $\frac{25}{256}$

Total number of ways in which $4$ tickets can be drawn $5$ times $=4^5=1024.$
Favourable number of ways of getting sum $23$ is
$=$ cofficient of $x^{23}$ in $(x^{00}+x^{01}+x^{10}+x^{11}{)}^5$
Let us solve the binomial expansion
$(x^{00}+x^{01}+x^{10}+x^{11}{)}^5$
$={\left(\right(1+x\left)\right(1+x^{10}\left)\right)}^5=(1+x{)}^5(1+x^{10}{)}^5$
$=(1+5x+10x^2+10x^3+5x^4+x^5)(1+5x^{10}+10x^{20}+\cdots )$
$\therefore$ cofficient of $x^{23}$ is $100.$
$\Rightarrow$ Required probability $=\frac{100}{1024}=\frac{25}{256}$