Probability of Picking 4 Specific Values out of 6
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Say you have six values (V1, V2, V3, V4, V5, V6). Four of the values are "1", the other two are "0". You randomly pick exactly five of them. What are the odds that you pick all four 1's and one 0?
An example set could be [1 0 0 1 1 1]
My best guess would be 1/6 chance, based on a C(6,5) combination, but I'm not 100% sure that's the correct way to approach this problem.
probability
asked Jan 23 at 16:02
EgerStuEgerStu
32
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add a comment |
$begingroup$
Say you have six values (V1, V2, V3, V4, V5, V6). Four of the values are "1", the other two are "0". You randomly pick exactly five of them. What are the odds that you pick all four 1's and one 0?
An example set could be [1 0 0 1 1 1]
My best guess would be 1/6 chance, based on a C(6,5) combination, but I'm not 100% sure that's the correct way to approach this problem.
probability
asked Jan 23 at 16:02
EgerStuEgerStu
32
$endgroup$
$begingroup$
Thanks for reminding me, I've added that.
$endgroup$
– EgerStu
Jan 23 at 16:12
add a comment |
$begingroup$
Say you have six values (V1, V2, V3, V4, V5, V6). Four of the values are "1", the other two are "0". You randomly pick exactly five of them. What are the odds that you pick all four 1's and one 0?
An example set could be [1 0 0 1 1 1]
My best guess would be 1/6 chance, based on a C(6,5) combination, but I'm not 100% sure that's the correct way to approach this problem.
probability
asked Jan 23 at 16:02
EgerStuEgerStu
32
$endgroup$
Say you have six values (V1, V2, V3, V4, V5, V6). Four of the values are "1", the other two are "0". You randomly pick exactly five of them. What are the odds that you pick all four 1's and one 0?
An example set could be [1 0 0 1 1 1]
My best guess would be 1/6 chance, based on a C(6,5) combination, but I'm not 100% sure that's the correct way to approach this problem.
probability
probability
asked Jan 23 at 16:02
EgerStuEgerStu
32
asked Jan 23 at 16:02
EgerStuEgerStu
32
edited Jan 23 at 16:12
EgerStu
asked Jan 23 at 16:02
EgerStuEgerStu
32
asked Jan 23 at 16:02
EgerStuEgerStu
32
asked Jan 23 at 16:02
EgerStuEgerStu
32
32
$begingroup$
Thanks for reminding me, I've added that.
$endgroup$
– EgerStu
Jan 23 at 16:12
add a comment |
$begingroup$
Thanks for reminding me, I've added that.
$endgroup$
– EgerStu
Jan 23 at 16:12
$begingroup$
Thanks for reminding me, I've added that.
$endgroup$
– EgerStu
Jan 23 at 16:12
$begingroup$
Thanks for reminding me, I've added that.
$endgroup$
– EgerStu
Jan 23 at 16:12
add a comment |
1 Answer
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There are six equiprobable ways to choose five values (just consider the one you leave out), but two of them are admissible (you leave out one zero). The probability is therefore $frac26=frac13$
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
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$begingroup$
There are six equiprobable ways to choose five values (just consider the one you leave out), but two of them are admissible (you leave out one zero). The probability is therefore $frac26=frac13$
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
$endgroup$
add a comment |
$begingroup$
There are six equiprobable ways to choose five values (just consider the one you leave out), but two of them are admissible (you leave out one zero). The probability is therefore $frac26=frac13$
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
$endgroup$
add a comment |
$begingroup$
There are six equiprobable ways to choose five values (just consider the one you leave out), but two of them are admissible (you leave out one zero). The probability is therefore $frac26=frac13$
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
$endgroup$
There are six equiprobable ways to choose five values (just consider the one you leave out), but two of them are admissible (you leave out one zero). The probability is therefore $frac26=frac13$
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
answered Jan 23 at 16:15
saulspatzsaulspatz
16.1k31331
16.1k31331
add a comment |
add a comment |
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$begingroup$
Thanks for reminding me, I've added that.
$endgroup$
– EgerStu
Jan 23 at 16:12