Question

A bag contains 7 tickets marked with the numbers 0, 1, 2, 3, 4, 5, 6, respectively. A ticket is drawn & replaced. Then the chance that after 4 drawings, the sum of the numbers drawn is 8 is

• A. 165/2401
• B. 149/2401
• C. 3/49
• D. none

Answer 1

The correct option is B 149/2401

In four drawings, the no. of possible outcomes is $7^4=2401.$

We want to find the number of all possible outcomes such that the sum of four no's drawn is $8$ and this no. is equal to the coefficient of $x^8$ in the expansion of

${(x^0+x^1+x^2+x^3+x^4+x^5+x^6)}^4$

$={\left(\frac{1-x^7}{1-x}\right)}^4={(1-x^7)}^4{(1-x)}^{-4}$

$=(1-{{}^{4}C}_1{x}^7+{{}^{4}C}_2{x}^{14}-{{}^{4}C}_3{x}^{21}+{{}^{4}C}_4{x}^{28})\times (a_0+a_{1}x+a_2{x}^2+....)$

Where $a_r$ is the coefficient of $x^r$ in ${(1-x)}^{-4}$

Hence, the no. of favorable outcomes.

$\Rightarrow a_8-4a_1=\frac{\mathrm{4.5.6....}(8+3)}{8!}-4\times 4$

$(\therefore$ In ${(1-x)}^{-4},T_{r+1}$

$=\frac{(-4)(-5)(-6).....(-4-r+1)}{r}{(-x)}^r$

$\Rightarrow a^r=\frac{\mathrm{4.5.6.....}(r+3)}{r!}$

Hence, number of favorable outcomes

$=\frac{\mathrm{4.5.6.7.8.9.10.11}}{\mathrm{1.2.3.4.5.6.7.8}}-16$

$=165-16$

$=149$

$\therefore$ Required probability $=\frac{149}{2401}.$

Hence, the answer is $\frac{149}{2401}.$