
From a pack of 52 cards, two cards are drawn one by one without replacement. The probability that first drawn card is a king and second is a queen, is
A.\[ \dfrac{2}{13}\]
B. \[ \dfrac{8}{663}\]
C. \[ \dfrac{4}{663}\]
D. \[ \dfrac{103}{663}\]
Answer
161.1k+ views
Hint: We all have a basic idea of cards. There are 52 cards in a pack. Among them, there are 4 types of cards such as 13 black spades, 13 red hearts, 13 blue diamonds and 13 green clubs.
Formula Used: Let E be an event then probability of E = P(E)= Number of favourable outcomes of E/Total number of outcomes of E.
Complete step by step solution:We all have a basic idea of cards. There are 52 cards in a pack. Among them, there are 4 types of cards such as 13 black spades, 13 red hearts, 13 blue diamonds and 13 green clubs. There are 4 king cards and 4 queen cards.
If we draw one king card among 52 cards one king card can be drawn in \(\dfrac{4}{{52}}\) ways and for each of drawn, one queen card can be drawn in \(\dfrac{4}{{51}}\)as one card has been drawn already (52-1=51, but there are 4 queen cards) .
Therefore, the probability that first drawn card is a king and second is a queen = \(\dfrac{4}{{52}}.\dfrac{4}{{51}} = \dfrac{4}{{663}}\) .
Option ‘C’ is correct
Note: This type of problem is very easy if you have the knowledge of the definition of probability and basic idea of pack of cards.
Formula Used: Let E be an event then probability of E = P(E)= Number of favourable outcomes of E/Total number of outcomes of E.
Complete step by step solution:We all have a basic idea of cards. There are 52 cards in a pack. Among them, there are 4 types of cards such as 13 black spades, 13 red hearts, 13 blue diamonds and 13 green clubs. There are 4 king cards and 4 queen cards.
If we draw one king card among 52 cards one king card can be drawn in \(\dfrac{4}{{52}}\) ways and for each of drawn, one queen card can be drawn in \(\dfrac{4}{{51}}\)as one card has been drawn already (52-1=51, but there are 4 queen cards) .
Therefore, the probability that first drawn card is a king and second is a queen = \(\dfrac{4}{{52}}.\dfrac{4}{{51}} = \dfrac{4}{{663}}\) .
Option ‘C’ is correct
Note: This type of problem is very easy if you have the knowledge of the definition of probability and basic idea of pack of cards.
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