Probability of obtaining at most one negative value in $5$ trials












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In an experiment, positive and negative values are equally likely to occur. What is the probability of obtaining at most one negative value in five trials?



$$P(text{obtaining at most one negative value in $5$ trials}) = ^5C_1 cdot left(frac{1}{2}right) cdot left(frac{1}{2}right)^4 = frac{5}{32}$$
But ans is $frac 6{32}$.










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    0














    In an experiment, positive and negative values are equally likely to occur. What is the probability of obtaining at most one negative value in five trials?



    $$P(text{obtaining at most one negative value in $5$ trials}) = ^5C_1 cdot left(frac{1}{2}right) cdot left(frac{1}{2}right)^4 = frac{5}{32}$$
    But ans is $frac 6{32}$.










    share|cite|improve this question



























      0












      0








      0







      In an experiment, positive and negative values are equally likely to occur. What is the probability of obtaining at most one negative value in five trials?



      $$P(text{obtaining at most one negative value in $5$ trials}) = ^5C_1 cdot left(frac{1}{2}right) cdot left(frac{1}{2}right)^4 = frac{5}{32}$$
      But ans is $frac 6{32}$.










      share|cite|improve this question















      In an experiment, positive and negative values are equally likely to occur. What is the probability of obtaining at most one negative value in five trials?



      $$P(text{obtaining at most one negative value in $5$ trials}) = ^5C_1 cdot left(frac{1}{2}right) cdot left(frac{1}{2}right)^4 = frac{5}{32}$$
      But ans is $frac 6{32}$.







      probability






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      edited Jan 4 at 16:41









      amWhy

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      asked Jul 6 '16 at 10:02









      akashakash

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          You calculated the probability of getting exactly one negative. It says "at most one", so you have to add in the probability of getting zero negatives, which is $1/32$.
          (also, your $36$s should be $32$s)






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            You calculated the probability of getting exactly one negative. It says "at most one", so you have to add in the probability of getting zero negatives, which is $1/32$.
            (also, your $36$s should be $32$s)






            share|cite|improve this answer


























              1














              You calculated the probability of getting exactly one negative. It says "at most one", so you have to add in the probability of getting zero negatives, which is $1/32$.
              (also, your $36$s should be $32$s)






              share|cite|improve this answer
























                1












                1








                1






                You calculated the probability of getting exactly one negative. It says "at most one", so you have to add in the probability of getting zero negatives, which is $1/32$.
                (also, your $36$s should be $32$s)






                share|cite|improve this answer












                You calculated the probability of getting exactly one negative. It says "at most one", so you have to add in the probability of getting zero negatives, which is $1/32$.
                (also, your $36$s should be $32$s)







                share|cite|improve this answer












                share|cite|improve this answer



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                answered Jul 6 '16 at 10:04









                florenceflorence

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