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(B) The total number of ways to choose $2$ cards from the same suit is given by choosing one suit out of $4$ and then choosing $2$ cards out of $13$ f

Vedclass · Accepted Answer

$\frac{2}{13}$

Vedclass · Answer

(B) The total number of ways to choose $2$ cards from the same suit is given by choosing one suit out of $4$ and then choosing $2$ cards out of $13$ from that suit. Total ways $= 4 	imes \binom{13}{2} = 4 	imes \frac{13 	imes 12}{2} = 312$. Alternatively, since the condition is that the cards are from the same suit, we consider the sample space where both cards belong to the same suit. There are $4$ suits, and for each suit, there are $\binom{13}{2} = 78$ ways to pick $2$ cards. Total outcomes $= 4 	imes 78 = 312$. In each suit, the prime-numbered cards are ${2, 3, 5, 7}$ (total $4$ cards) and the face cards are ${J, Q, K}$ (total $3$ cards). We need one face card and one prime-numbered card from the same suit. For one specific suit, the number of ways to choose one face card and one prime card is $3 	imes 4 = 12$. Since there are $4$ suits, the total number of favourable outcomes is $4 	imes (3 	imes 4) = 48$. However, the question implies the probability given that they are from the same suit. Given the cards are from the same suit, the total ways to pick $2$ cards is $\binom{13}{2} = 78$. The number of ways to pick one face card ($3$ options) and one prime card ($4$ options) is $3 	imes 4 = 12$. Since the order does not matter, we have $12$ ways. Probability $= \frac{12}{78} = \frac{2}{13}$.

If $2$ cards drawn at random from a well-shuffled pack of $52$ playing cards are from the same suit, then the probability of getting a face card and a card having a prime number is

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