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anonymous

  • one year ago

what is the probability of getting an odd number on a fair die and a tail on a coin flip

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  1. anonymous
    • one year ago
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    @johnweldon1993

  2. johnweldon1993
    • one year ago
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    Hey! So first...a fair die has 6 sides with an equal probability for each right? so the probability for any 1 number would be \(\large \dfrac16\) And a coin has 2 sides...with equal probability again...so that chance would be \(\large \dfrac12\) So how many odd numbers on are a die?

  3. anonymous
    • one year ago
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    3

  4. johnweldon1993
    • one year ago
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    Right...so there are 3 out of 6 numbers on a die that are off....so the chance of getting any random odd number on a die is 3/6 or 1/2 right? SO!! The probability of getting an odd on a die is \(\large \dfrac12\) and the probability of getting a tails on a coin is also \(\large \dfrac12\) so the probability we want for those together is \[\large \frac{1}{2} \times \frac{1}{2} =?\]

  5. johnweldon1993
    • one year ago
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    btw, odd* not off* I'm sure you know that but wanted to clarify :)

  6. anonymous
    • one year ago
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    i have another onnneee

  7. anonymous
    • one year ago
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    what is the probability of getting a number less than three on a fair die and a tail on a coin flip

  8. johnweldon1993
    • one year ago
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    Well we already know the tails probability is that 1/2 So...a number LESS THAN 3 how many numbers are less than 3 on a die?

  9. anonymous
    • one year ago
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    2

  10. johnweldon1993
    • one year ago
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    Indeed, so that probability is 2 numbers out of 6 ...so 2/6 or 1/3 so our total probability of the 2 events is \(\large \frac{1}{3} \times \frac{1}{2} =?\)

  11. anonymous
    • one year ago
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    what is the probability of choosing a 2 and a face card from a deck of cards

  12. johnweldon1993
    • one year ago
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    Okay...standard deck of cards has 52 cards right? How many 2's are in there? and how many face cards are in there?

  13. anonymous
    • one year ago
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    4 2's and 12 face cards i believe

  14. johnweldon1993
    • one year ago
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    mmhmm :) 4 suits and 1 2 in each so there are 4 2's and 3 face cards in each suit so 12 face cards SO! Now this is where I'm a little confused...if we are choosing both cards at the SAME time...we can consider these independent... however if we choose a 2 from the deck...and THEN choose a face card from the only 51 cards remaining...that changes things...so I'm not too sure how to proceed

  15. johnweldon1993
    • one year ago
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    But I think...since it doesnt SPECIFY one and THEN the other...I think its just the basic choose 2 random cards at the same time...so The probability of choosing a 2 out of 52 cards is \(\large \dfrac2{52} = \dfrac{1}{26}\) And the probability of choosing a face card would be \(\large \dfrac{12}{52} = \dfrac{3}{13}\) So the total probability would be \(\large \frac{1}{26} \times \frac{3}{13} = ?\)

  16. anonymous
    • one year ago
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    3/338 which is 0.00887573964 right so the answer would be 0.00?

  17. johnweldon1993
    • one year ago
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    Omg....brain fart -_- Sorry...probability of choosing a 2 would be \[\large \frac{4}{52} = \frac{1}{13}\] sorry about that -_- lol so its actually \[\large \frac{1}{13} \times \frac{3}{13}\]

  18. anonymous
    • one year ago
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    3/169 which is 0.0177514793

  19. johnweldon1993
    • one year ago
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    There you go...that's better lol as soon as I saw you put 0.00 I was like...well THAT means impossible! lol sorry about that!

  20. anonymous
    • one year ago
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    lol its all good

  21. anonymous
    • one year ago
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    but my options are 0.00, 0.077, 0.25, 0.3077

  22. johnweldon1993
    • one year ago
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    Argh see this is why I hate probability questions that are NOT specific lol Okay....I'm going to sum up here based on what I think \[\large P(2) = \frac{4}{52} = \frac{1}{13}\] \[\large P(face) = \frac{12}{52} = \frac{3}{13}\] \[\large P(together) = P(2) + P(face) = \frac{1}{13} + \frac{3}{13} = \frac{4}{13} = 0.3077\]

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