Question

Four tickets marked $00,01,10$ and $11$, 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 numbers on the tickets is $15$ is

• A. $\frac{3}{1024}$
• B. $\frac{5}{1024}$
• C. $\frac{7}{1024}$
• D. None of these

Answer 1

Answer: (B)

$\frac{5}{1024}$

Let $S$ be the sample space and $E$ be the required event.

Now, $n\left(S\right)=$ total number of cases $=4^5=1024.$

and, $n\left(E\right)=$coefficient of $x^{15}$ in ${(x^0+x+x^{10}+x^{11})}^5$

$=$coefficient of $x^{15}$ in $\left[{(1+x)}^5{(1+x^{10})}^5\right]$

$=$coefficient of $x^{15}$in $(1+5x+10x^2+10x^3+5x^4+x^5)\times (1+5x^{10}+10x^{20}+...)=5$

$\therefore$ Required probability, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{5}{1024}.$