Two cards are drawn from a pack of 52 playing | vedclass

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(A) Case $1$: With replacement.$p_1 = P(	ext{First is Queen}) 	imes P(	ext{Second is Diamond}) = \frac{4}{52} 	imes \frac{13}{52} = \frac{1}{13} \

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(A) Case $1$: With replacement.$p_1 = P(	ext{First is Queen}) 	imes P(	ext{Second is Diamond}) = \frac{4}{52} 	imes \frac{13}{52} = \frac{1}{13} 	imes \frac{1}{4} = \frac{1}{52}$.Case $2$: Without replacement.Let $Q_1$ be the event that the first card is a queen and $D_2$ be the event that the second card is a diamond.If the first card is the Queen of Diamonds, $P(Q_1 \cap D_2) = P(Q_1 \cap D_1) 	imes P(D_2 | Q_1 \cap D_1) = \frac{1}{52} 	imes \frac{12}{51}$.If the first card is a Queen other than the Queen of Diamonds, $P(Q_1 \cap D_2) = P(Q_1 \cap D_1^c) 	imes P(D_2 | Q_1 \cap D_1^c) = \frac{3}{52} 	imes \frac{13}{51}$.$p_2 = \frac{1 	imes 12 + 3 	imes 13}{52 	imes 51} = \frac{12 + 39}{52 	imes 51} = \frac{51}{52 	imes 51} = \frac{1}{52}$.Therefore, $\frac{p_1}{p_2} = \frac{1/52}{1/52} = 1$.

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