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Johnmathstudent

  • 2 years ago

If I flip four coins what is the probability that at least two of the coins are heads?

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  1. Johnmathstudent
    • 2 years ago
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    Please help

  2. amistre64
    • 2 years ago
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    its the opposite of P(at most 1 is heads)

  3. amistre64
    • 2 years ago
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    1 - P(0) - P(1)

  4. MAYSTEPH
    • 2 years ago
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    do you have choices

  5. amistre64
    • 2 years ago
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    there are 16 outcomes tttt ttth ttht thtt httt ... 1 4 6 4 1 = 16 0 heads is 1 out of 16 1 head is 4 out of 16

  6. Johnmathstudent
    • 2 years ago
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    What about at least 2 heads

  7. amistre64
    • 2 years ago
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    at least 2 heads means that you have 2, 3, or 4 heads P(0) + P(1) + P(2) + P(3) + P(4) = 1 P(2) + P(3) + P(4) = 1 - P(0) - P(1)

  8. amistre64
    • 2 years ago
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    since P(0) = 1/16 and P(1) = 4/16 the solution is determined

  9. Johnmathstudent
    • 2 years ago
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    So what would the answer be written as?

  10. amistre64
    • 2 years ago
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    im not going to do the subtraction for you ....

  11. amistre64
    • 2 years ago
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    you can tell me what you get ... and i can verify if you did it correctly

  12. MAYSTEPH
    • 2 years ago
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    bro your even confusing me.... amistre64

  13. amistre64
    • 2 years ago
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    im not a mind reader, so youll have to be more detailed in what is confusing :)

  14. MAYSTEPH
    • 2 years ago
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    nevermind .. the probability would be 50% John

  15. amistre64
    • 2 years ago
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    the probability is not 50%; for starters, a probability is a number between 0 and 1 ... not a percentage.

  16. amistre64
    • 2 years ago
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    and for another, 1 - P(0) - P(1) does not equal .5000 or 1/2

  17. MAYSTEPH
    • 2 years ago
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    he asked for a probability. if you flip 4 coins and the odds of 2 of the comming out heads is 2/4 and tht equals 50% and wtf is p

  18. MAYSTEPH
    • 2 years ago
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    its not rocket science he just wants a simple answer.. idk whats wrong with tht

  19. amistre64
    • 2 years ago
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    \[\binom{4}{0}.5^0.5^4+\binom{4}{1}.5^1.5^3+\binom{4}{2}.5^2.5^2+\binom{4}{3}.5^3.5^1+\binom{4}{4}.5^n.4^0=1\] \[(1).5^0.5^4+(4).5^1.5^3+(6).5^2.5^2+(4).5^3.5^1+(1).5^4.5^0=1\]

  20. amistre64
    • 2 years ago
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    P(E) is notation for "probability of an event"

  21. amistre64
    • 2 years ago
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    the last 3 terms are 4 choose 2, 4 choose 3, and 4 choose 4 which is what we want if we are to determine "at least 2" out of 4

  22. amistre64
    • 2 years ago
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    subtracting the first 2 terms from each side gives us the solution

  23. MAYSTEPH
    • 2 years ago
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    your making this way more complex than it already isnt lol if you know the damn answer then tell him and quit arguing with a degenerate!

  24. amistre64
    • 2 years ago
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    i suggest you tone down your language. and, we are not providing a free answering service, we are offering a way to learn the material.

  25. amistre64
    • 2 years ago
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    my first attempt was to simplify it .... since that did not work i resorted to the longer, more detailed version of it all.

  26. Johnmathstudent
    • 2 years ago
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    So the answer is .6875?

  27. Johnmathstudent
    • 2 years ago
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    Or 11/16

  28. amistre64
    • 2 years ago
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    11/16 is correct either way is fine, but the format is up to whoever is doing the grading.

  29. Johnmathstudent
    • 2 years ago
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    Thanks I just forgot how to do this from the beginning of the year

  30. amistre64
    • 2 years ago
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    you did fine :) good luck

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