Probability: Draw without replacement

Psuccess=nCr×pPr×(sp)P(nr)sPn P_{success}= \frac{ ^nC_r \times ^pP_r \times ^{(s-p)}P_{(n-r)}}{^sP_n}

=pCr×(sp)C(nr)sCn = \dfrac{ ^pC_r \times ^{(s - p)}C_{(n - r)} }{^sC_n}


nn=number of draws

pp=maximum number of successful draws

ss=maximum number of draws

rr= number of successful draws obtained

(still incomplete) It's similar to the probability of Bernoulli trial

Bernoulli trial: either pass or fail, but with replacement

Psuccess= nCr×pr×(1p)nrP_{success} = \text{ }^nC_r \times p^r \times (1-p)^{n-r}


nCr represents the total number of configurations of success and failure from the obtained ones^nC_r \text{ represents the total number of configurations of success and failure from the obtained ones}

pPr represents the total number of successful choices ^pP_r \text{ represents the total number of successful choices}

(sp)P(nr) represents the total number of unsuccessful choices ^{(s-p)}P_{(n-r)} \text{ represents the total number of unsuccessful choices}

sPn represents the total number of combinations that can be obtained for n draws ^sP_n \text{ represents the total number of combinations that can be obtained for } n \text{ draws}

This is also called a hypergeometric distribution

#Combinatorics

Note by Lin Shun Hao
1 year, 4 months ago

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