Probability of obtaining at most one negative value in $5$ trials
$begingroup$
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
edited Jan 4 at 16:41
amWhy
1
asked Jul 6 '16 at 10:02
akashakash
7217
$endgroup$
add a comment |
$begingroup$
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
edited Jan 4 at 16:41
amWhy
1
asked Jul 6 '16 at 10:02
akashakash
7217
$endgroup$
add a comment |
$begingroup$
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
edited Jan 4 at 16:41
amWhy
1
asked Jul 6 '16 at 10:02
akashakash
7217
$endgroup$
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
probability
edited Jan 4 at 16:41
amWhy
1
asked Jul 6 '16 at 10:02
akashakash
7217
edited Jan 4 at 16:41
amWhy
1
asked Jul 6 '16 at 10:02
akashakash
7217
edited Jan 4 at 16:41
amWhy
1
edited Jan 4 at 16:41
amWhy
1
edited Jan 4 at 16:41
amWhy
1
1
asked Jul 6 '16 at 10:02
akashakash
7217
asked Jul 6 '16 at 10:02
akashakash
7217
asked Jul 6 '16 at 10:02
akashakash
7217
7217
add a comment |
add a comment |
1 Answer
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$begingroup$
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)
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
$endgroup$
add a comment |
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$begingroup$
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)
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
$endgroup$
add a comment |
$begingroup$
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)
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
$endgroup$
add a comment |
$begingroup$
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)
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
$endgroup$
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)
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
answered Jul 6 '16 at 10:04
florenceflorence
11.5k12045
11.5k12045
add a comment |
add a comment |
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