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GOODMAN Group Title

Two cards are drawn from a standard deck of 52 cards without replacement. What is the probability that both cards are greater than 2 and less than 9?

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    Okay, I'll help you with this :)

    • 2 years ago
  2. GOODMAN Group Title
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    Thanks little bro, lol. I am studying and forgot this all.

    • 2 years ago
  3. ParthKohli Group Title
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    What would be the probability of getting a card between 2 and 9 in the first pick?

    • 2 years ago
  4. GOODMAN Group Title
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    2/52 and 9/52

    • 2 years ago
  5. ParthKohli Group Title
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    No...it's saying that it's greater than 2 and less than 9. 3,4,5,6,7,8 are the numbers that qualify for me :)

    • 2 years ago
  6. GOODMAN Group Title
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    So we have to get the ones in between?

    • 2 years ago
  7. ParthKohli Group Title
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    Yes.. exactly

    • 2 years ago
  8. ParthKohli Group Title
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    Since there are 3,4,5,6,7,8 in four suits, the probability would be: \( \color{Black}{\Rightarrow \Large {4 \times 6 \over 52} }\) For the first pick.

    • 2 years ago
  9. ParthKohli Group Title
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    Simplifying further: \( \color{Black}{\Rightarrow \Large {24 \over 52} = {12 \over 26} = {6 \over 13} }\)

    • 2 years ago
  10. ParthKohli Group Title
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    Now we've picked the first. We have 1 card LESS in the deck because we haven't replaced the cards.

    • 2 years ago
  11. ParthKohli Group Title
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    So this time the denominator of the probability would become 51. The numerator will also get one less because we have assumed that we have picked one card which is 3,4,5,6,7 or 8.

    • 2 years ago
  12. ParthKohli Group Title
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    There are 23 cards left that we want. 51 total cards left. \( \color{Black}{\Rightarrow \Large {23 \over 51} }\)

    • 2 years ago
  13. GOODMAN Group Title
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    Okay...so since we had not replaced, we have to subtract 1?

    • 2 years ago
  14. ParthKohli Group Title
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    Now if we want two things to happen at the same time, we shall multiply the probabilities. \( \color{Black}{\Rightarrow \Large {13 \over 26} \times {23 \over 51}}\)

    • 2 years ago
  15. ParthKohli Group Title
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    We have to subtract 1 from both numerator and denominator.

    • 2 years ago
  16. GOODMAN Group Title
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    Thats it?

    • 2 years ago
  17. ParthKohli Group Title
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    Nope

    • 2 years ago
  18. GOODMAN Group Title
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    Wait. so after we multiply, we have to simplify further too?

    • 2 years ago
  19. ParthKohli Group Title
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    Oops.. I meant this: \( \color{Black}{\Rightarrow \Large {6 \over 13} \times {23 \over 51}}\)

    • 2 years ago
  20. ParthKohli Group Title
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    Multiply the fractions. The fractions are in their simplest forms so when we'll multiply it'd be in the simplest form.

    • 2 years ago
  21. GOODMAN Group Title
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    Yes, i did that. Okay, so 46/221 is final answer?

    • 2 years ago
  22. ParthKohli Group Title
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    How did you get that?

    • 2 years ago
  23. ParthKohli Group Title
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    Oh yes

    • 2 years ago
  24. ParthKohli Group Title
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    \( \color{Black}{\Rightarrow \Large {2 \over 13} \times {23 \over 17} }\)

    • 2 years ago
  25. ParthKohli Group Title
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    Correct! :D

    • 2 years ago
  26. GOODMAN Group Title
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    Okay, thats awesome!! Thanks soo much Parth!! Probability is my weak spot :/

    • 2 years ago
  27. ParthKohli Group Title
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    Haha probability is easy...getting the hang of what it involves is important

    • 2 years ago
  28. GOODMAN Group Title
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    Yea, see, im reviewing this since last semester, lol.

    • 2 years ago
  29. ParthKohli Group Title
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    And I've helped someone on this site after a long time. we usually just answer questions ://

    • 2 years ago
  30. GOODMAN Group Title
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    Thanks, I actually learned :D

    • 2 years ago
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