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GOODMAN

  • 2 years ago

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?

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  1. ParthKohli
    • 2 years ago
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    Okay, I'll help you with this :)

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

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

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

  5. ParthKohli
    • 2 years ago
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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 :)

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

  7. ParthKohli
    • 2 years ago
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    Yes.. exactly

  8. ParthKohli
    • 2 years ago
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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.

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

  10. ParthKohli
    • 2 years ago
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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.

  11. ParthKohli
    • 2 years ago
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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.

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

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

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

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

  16. GOODMAN
    • 2 years ago
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    Thats it?

  17. ParthKohli
    • 2 years ago
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    Nope

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

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

  20. ParthKohli
    • 2 years ago
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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.

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

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

  23. ParthKohli
    • 2 years ago
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    Oh yes

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

  25. ParthKohli
    • 2 years ago
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    Correct! :D

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

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

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

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

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

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