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(C) The probability that exactly $5$ cards are drawn before the first ace appears means the first $5$ cards are non-aces and the $6$th card is an ace.

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) The probability that exactly $5$ cards are drawn before the first ace appears means the first $5$ cards are non-aces and the $6$th card is an ace.$P = \frac{48}{52} 	imes \frac{47}{51} 	imes \frac{46}{50} 	imes \frac{45}{49} 	imes \frac{44}{48} 	imes \frac{4}{47}$Simplifying the expression:$P = \frac{4}{49} 	imes \left( \frac{48 	imes 47 	imes 46 	imes 45 	imes 44}{52 	imes 51 	imes 50 	imes 48 	imes 47} ight) = \frac{4}{49} 	imes \left( \frac{46 	imes 45 	imes 44}{52 	imes 51 	imes 50} ight)$$P = \frac{4}{49} 	imes \left( \frac{23 	imes 2 	imes 9 	imes 5 	imes 11 	imes 4}{13 	imes 4 	imes 17 	imes 3 	imes 25 	imes 2} ight) = \frac{4}{49} 	imes \left( \frac{23 	imes 3 	imes 11}{13 	imes 17 	imes 5} ight)$Here, the prime factors are $p_1=23, p_2=11, p_3=3$ and $p_4=13, p_5=17, p_6=5$.Thus, $\max \{p_i\} = 23$ and $\min \{p_i\} = 3$.Therefore, $\max \{p_i\} - \min \{p_i\} = 23 - 3 = 20$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

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List-$I$ List-$II$
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