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anonymous

  • 5 years ago

You have a 3-card deck containing a king, a queen, and a jack. You draw a random card, then put it back and draw a second random card. Calculate the probability that you draw exactly 1 king.

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  1. anonymous
    • 5 years ago
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    Same as the earlier question about the jack: 2/3*1/2 = 1/3

  2. dumbcow
    • 5 years ago
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    P(draw both jacks) = 1/3*1/3 = 1/9 P(no jacks) = 2/3*2/3 = 4/9 P(1 jack) = 1 - 1/9 - 4/9 = 4/9

  3. dumbcow
    • 5 years ago
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    its different because there is replacement

  4. anonymous
    • 5 years ago
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    Probability of taking out one king in one attempt is 1/3 Correspondingly, prob of 'not' taking out a king in another attempt is 2/3 So the total probability of taking out 'just one' king is 2/9!

  5. dumbcow
    • 5 years ago
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    what if you pick the king on the 2nd attempt, not on the first

  6. anonymous
    • 5 years ago
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    yes! i pick it on 'any' one of the attempt! both the probs have the same sample set, so it shudnt matter

  7. dumbcow
    • 5 years ago
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    yes both situations have prob of 2/9 2/9 + 2/9 = 4/9

  8. anonymous
    • 5 years ago
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    the right answer is 4/9

  9. anonymous
    • 5 years ago
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    Getting king from first attempt only = 1/3 * 2/3 = 2/9 Getting king from second attempt = 2/3 * 1/3 = 2/9 Add them up , u get 4/9

  10. anonymous
    • 5 years ago
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    @dumbcow: I was thinking whether both are same or different, but yeah, you're right as there is actual time difference, sry!

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