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anonymous

  • one year ago

John’s piggy bank contained 7 dimes and 3 quarters. He pulled out 1 coin without looking. Without replacing the first coin, John then pulled out a second coin. What is the probability that both coins were dimes? Express your answer in simplest form.

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  1. anonymous
    • one year ago
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    My thought is that it's as simple as 2/10. Im not quite sure though.

  2. anonymous
    • one year ago
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    We have two steps for this. I'll help you through them one-by-one. The first step is figuring out what the probability is for pulling the first dime out. Which is represented as \[\large \frac{no.~of~dimes}{total~no.~of~coins}\]

  3. anonymous
    • one year ago
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    7/10 then I would presume?

  4. anonymous
    • one year ago
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    Yep. So our next step (3 in total, not 2 sorry) is to figure out what the probability of getting a dime would be if there is one less dime and one less coin. Which is represented as \[\large \frac{no.~of~coins~-1}{total~no.~of~coins~-1}\]

  5. anonymous
    • one year ago
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    So 6/9. With the reduced dime he would have presumably pulled, it takes one away from the amount of dimes, which in return reduces the amount of coins altogether.

  6. anonymous
    • one year ago
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    Right, good job. Before we move on to the final step, do you know what 6/9 will simplify to?

  7. anonymous
    • one year ago
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    2/3 of course.

  8. anonymous
    • one year ago
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    Great! So our final step is to multiply the two fractions together, to get the probability of both happening. So \[\large \frac{7}{10} \times \frac{2}{3}=~?\]

  9. anonymous
    • one year ago
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    14/30, which would reduce to 7/15. 7 is a prime number so that must be my answer.

  10. anonymous
    • one year ago
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    Cool! Thanks a lot. I appreciate your help!

  11. anonymous
    • one year ago
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    Right, it would be 7/15. And also, \[\huge \color{aqua}N\color{fuchsia}o \space \color{lime}P \color{orange}r \color{blue}o \color{maroon}b \color{red}l \color{olive}e \color{purple}m \ddot\smile \]

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