If two cards are drawn at random simultaneous | vedclass

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(C) In a deck of $52$ cards,each suit contains numbers $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$. Composite numbers are $\{4, 6, 8, 9, 10\}$ ($5$ cards

Vedclass · Accepted Answer

$\frac{102}{663}$

Vedclass · Answer

(C) In a deck of $52$ cards,each suit contains numbers $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$. Composite numbers are $\{4, 6, 8, 9, 10\}$ ($5$ cards per suit,total $20$). Multiples of $3$ are $\{3, 6, 9\}$ ($3$ cards per suit,total $12$). Let $C$ be the set of composite cards and $M$ be the set of multiples of $3$. $C = \{4, 6, 8, 9, 10\}$,$M = \{3, 6, 9\}$. Intersection $C \cap M = \{6, 9\}$ ($2$ cards per suit,total $8$). Cards in $C$ but not in $M$: $C \setminus M = \{4, 8, 10\}$ ($3$ cards per suit,total $12$). Cards in $M$ but not in $C$: $M \setminus C = \{3\}$ ($1$ card per suit,total $4$). We need one card from $C$ and one from $M$. Case $1$: One from $(C \setminus M)$ and one from $(M \setminus C) = 12 	imes 4 = 48$. Case $2$: One from $(C \setminus M)$ and one from $(C \cap M) = 12 	imes 8 = 96$. Case $3$: One from $(M \setminus C)$ and one from $(C \cap M) = 4 	imes 8 = 32$. Case $4$: Both from $(C \cap M) = \binom{8}{2} = 28$. Total favorable outcomes $= 48 + 96 + 32 + 28 = 204$. Total outcomes $= \binom{52}{2} = \frac{52 	imes 51}{2} = 1326$. Probability $= \frac{204}{1326} = \frac{102}{663}$.

If two cards are drawn at random simultaneously from a well-shuffled pack of $52$ playing cards,then the probability of getting a card having a composite number and a card having a number which is a multiple of $3$ is

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