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Question

(A) Let $E_1$ be the event of drawing an ace first and $E_2$ be the event of drawing a coloured card second.Total cards in a pack = $52$.Number of ace

Vedclass · Accepted Answer

$\frac{1}{26}$

Vedclass · Answer

(A) Let $E_1$ be the event of drawing an ace first and $E_2$ be the event of drawing a coloured card second.Total cards in a pack = $52$.Number of aces = $4$.$P(E_1) = \frac{4}{52} = \frac{1}{13}$.After drawing one ace,$51$ cards remain in the pack.There are $26$ coloured cards (hearts and diamonds) in a pack. If the first card drawn (an ace) was a red ace (heart or diamond),there are $25$ coloured cards left. If the first card was a black ace (spade or club),there are $26$ coloured cards left.However,in standard probability problems of this type,we consider the total coloured cards available. Since there are $2$ red aces and $2$ black aces:If $E_1$ is a red ace (prob $2/52$),$P(E_2|E_1) = 25/51$.If $E_1$ is a black ace (prob $2/52$),$P(E_2|E_1) = 26/51$.Total probability = $(2/52 	imes 25/51) + (2/52 	imes 26/51) = (2/52) 	imes (51/51) = 2/52 = 1/26$. Wait,re-evaluating the provided solution logic: The solution assumes $15$ coloured cards remain,which is incorrect. Let's re-calculate: Total coloured cards = $26$. Probability = $(2/52 	imes 25/51) + (2/52 	imes 26/51) = 1/26$. Given the options,let's check if 'coloured' implies something else. If the question implies $26$ coloured cards and we draw an ace,the probability is $1/26$. Since $1/26$ is option $A$,we select $A$.

Two cards are drawn one by one from a pack of cards. The probability of getting the first card as an ace and the second as a coloured card is (the first card is not replaced before drawing the second card).

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