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You have an ace, king, queen, and jack from a deck of cards. After shuffling the cards, you draw one card. Then you replace the card, shuffle again, and again draw a card. What is the probability that you will draw a king and a queen in either order? An ace, king, queen, and jack playing card. CLEARCHECK The total number of possible outcomes is x . The probability of drawing a king and a queen in either order is x

Sagot :

semsee45

Answer:

[tex]\dfrac{2}{169}[/tex]

Step-by-step explanation:

A standard 52-card deck comprises 4 suits (Spades, Hearts, Diamonds, and Clubs).  

Each suit comprises 10 numerical cards (numbered 2 through 10, plus an "ace") and 3 "court" cards (jack, queen and king).

Therefore:

  • The total number of possible outcomes is 52.
  • There are 4 kings and 4 queens in a standard deck of cards.

[tex]\boxed{\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}}[/tex]

[tex]\implies \sf P(Queen)=\dfrac{4}{52}=\dfrac{1}{13}[/tex]

[tex]\implies \sf P(King)=\dfrac{4}{52}=\dfrac{1}{13}[/tex]

Therefore, the probability of drawing a king, replacing the card, and drawing a queen is:

[tex]\implies \sf P(King)\;and\;P(Queen)=\dfrac{1}{13}\times \dfrac{1}{13}=\dfrac{1}{169}[/tex]

Similarly, the probability of drawing a queen, replacing the card, and drawing a king is:

[tex]\implies \sf P(Queen)\;and\;P(King)=\dfrac{1}{13}\times \dfrac{1}{13}=\dfrac{1}{169}[/tex]

Therefore, the probability of drawing a king then a queen, or a queen then a king is:

[tex]\implies \sf P(King\;and\;Queen)\;or\;P(Queen\;and\;King)=\dfrac{1}{169}+\dfrac{1}{169}=\dfrac{2}{169}[/tex]

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