Q78 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

A boy throws an unbiased die. Whenever he gets 1 on the die he has a further chance to throw it once again immediately. The probability that the boy gets a score of 7 in this process is

  • A
    $\frac{1}{5}\left(1-\frac{1}{6^5}\right)$
  • B
    $\frac{1}{30}\left(1-\frac{1}{6^4}\right)$
  • Correct answer
    C
    $\frac{1}{30}\left(1-\frac{1}{6^5}\right)$
  • D
    $\frac{1}{5}\left(1-\frac{1}{6^4}\right)$

Solution

$P($ any number $)=\frac{1}{6}$ Ways to get 7 : Probability $\begin{aligned} & {[1,6]: \frac{1}{6} \cdot \frac{1}{6}=\left(\frac{1}{6}\right)^2} \\ & {[1,1,5]: \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}=\left(\frac{1}{6}\right)^3} \\ & {[1,1,1,4]:\left(\frac{1}{6}\right)^3 \cdot \frac{1}{6}=\left(\frac{1}{6}\right)^4} \\ & {[1,1,1,1,3]:\left(\frac{1}{6}\right)^4\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^5} \\ & {[1,1,1,1,1,2]:\left(\frac{1}{6}\right)^5\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^6}\end{aligned}$ Total probability $\begin{aligned} & =\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^3+\ldots \ldots+\left(\frac{1}{6}\right)^6 \\ & =\left(\frac{1}{6}\right)^2 \frac{\left[1-\left(\frac{1}{6}\right)^5\right]}{\left[1-\frac{1}{6}\right]}=\frac{1}{30}\left(1-\frac{1}{6^5}\right.\end{aligned}$

Hence, the correct answer is option: 'C'.

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