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anonymous

  • 5 years ago

Ten students are applying for 3 positions on a team. the students include 4 boys (adam, alex, anthony, and arnold) and 6 girls (abbey, aurora, agnes, alice, amanda, and anna). all the students have an equal chance of being selected. find the probability that the students selected will include: a) 3 girls b) 1 boy and 2 girls c) at most 1 girl d) adam, anthony, and alice e) agnes and 2 other students

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  1. amistre64
    • 5 years ago
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    6 in 10 that it has 1 girl.. ithink :)

  2. anonymous
    • 5 years ago
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    This is a fundamental counting princple problem: (a) we need all three to be girls. The number of ways we can select three girls is 6 * 5 * 4 (since there is no replacement). The number of all selections is 10 * 9 * 8. Thus, P(3 girls) = (6*5*4)/(10*9*8)

  3. amistre64
    • 5 years ago
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    no replacement accounts for the duplicates right?

  4. anonymous
    • 5 years ago
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    yea

  5. anonymous
    • 5 years ago
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    (b) 1 boy and 2 girls is similar to above. Since order is not important, we have 4 * 6 * 5, 4 boys, then 6 girls and then 5 girls. The probability of all is still 10*9*8. (a) duplicates? yea, it means after we pick a girl we don't pick her again.

  6. anonymous
    • 5 years ago
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    can u do it using permutations and combinations please

  7. anonymous
    • 5 years ago
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    No. I know they teach it that way, but I think that is a bad way to go. Understanding it this way helps you solve every question, not just the ones in the form you're familiar with.

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