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(C) The probability of getting any specific number on a fair die is $P = \frac{1}{6}$.The boy gets a total score of $7$ if the sequence of outcomes su

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$\frac{1}{30}\left(1-\frac{1}{6^5}\right)$

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(C) The probability of getting any specific number on a fair die is $P = \frac{1}{6}$.The boy gets a total score of $7$ if the sequence of outcomes sums to $7$. Since he gets an extra throw only if he rolls a $1$, the possible sequences are:$1. [1, 6]: P = \frac{1}{6} 	imes \frac{1}{6} = \left(\frac{1}{6}ight)^2$$2. [1, 1, 5]: P = \frac{1}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \left(\frac{1}{6}ight)^3$$3. [1, 1, 1, 4]: P = \left(\frac{1}{6}ight)^3 	imes \frac{1}{6} = \left(\frac{1}{6}ight)^4$$4. [1, 1, 1, 1, 3]: P = \left(\frac{1}{6}ight)^4 	imes \frac{1}{6} = \left(\frac{1}{6}ight)^5$$5. [1, 1, 1, 1, 1, 2]: P = \left(\frac{1}{6}ight)^5 	imes \frac{1}{6} = \left(\frac{1}{6}ight)^6$The total probability is the sum of these independent sequences:$P(	ext{Total} = 7) = \left(\frac{1}{6}ight)^2 + \left(\frac{1}{6}ight)^3 + \left(\frac{1}{6}ight)^4 + \left(\frac{1}{6}ight)^5 + \left(\frac{1}{6}ight)^6$This is a geometric series with $a = \left(\frac{1}{6}ight)^2$, $r = \frac{1}{6}$, and $n = 5$ terms.Sum $= a \frac{1-r^n}{1-r} = \left(\frac{1}{6}ight)^2 \frac{1-(1/6)^5}{1-1/6} = \frac{1}{36} 	imes \frac{1-(1/6)^5}{5/6} = \frac{1}{36} 	imes \frac{6}{5} 	imes \left(1-\frac{1}{6^5}ight) = \frac{1}{30} \left(1-\frac{1}{6^5}ight)$.

List-$I$	List-$II$
$A. P(E_2) =$	$I. 2/3$
$B. P(E_1 \| E_2) =$	$II. 5/6$
$C. P(\bar{E}_2 \| E_1) =$	$III. 1/3$
$D. P(\bar{E}_1 \cup \bar{E}_2) =$	$IV. 1/2$

$A$ 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

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