Cards are drawn one by one without replacemen | vedclass

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Question

(B) There are $4$ aces and $48$ non-ace cards in a pack of $52$ cards.We want the first $10$ cards to be non-aces and the $11^{th}$ card to be an ace.

Vedclass · Accepted Answer

$\frac{164}{4165}$

Vedclass · Answer

(B) There are $4$ aces and $48$ non-ace cards in a pack of $52$ cards.We want the first $10$ cards to be non-aces and the $11^{th}$ card to be an ace.The probability is given by the product of probabilities of drawing $10$ non-aces followed by an ace:$P = \left( \frac{48}{52} 	imes \frac{47}{51} 	imes \frac{46}{50} 	imes \frac{45}{49} 	imes \frac{44}{48} 	imes \frac{43}{47} 	imes \frac{42}{46} 	imes \frac{41}{45} 	imes \frac{40}{44} 	imes \frac{39}{43} ight) 	imes \frac{4}{42}$Canceling common terms in the numerator and denominator:$P = \frac{48 	imes 47 	imes 46 	imes 45 	imes 44 	imes 43 	imes 42 	imes 41 	imes 40 	imes 39}{52 	imes 51 	imes 50 	imes 49 	imes 48 	imes 47 	imes 46 	imes 45 	imes 44 	imes 43} 	imes \frac{4}{42}$$P = \frac{42 	imes 41 	imes 40 	imes 39}{52 	imes 51 	imes 50 	imes 49} 	imes \frac{4}{42} = \frac{41 	imes 40 	imes 39}{52 	imes 51 	imes 50 	imes 49} 	imes 4$$P = \frac{41 	imes 4 	imes 39}{13 	imes 51 	imes 50 	imes 49} = \frac{164}{4165}$.

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is

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