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Picking Probabilities

Tool to make probabilities on picking objects. Calculation of probabilities of drawing objects (balls, beads, cards, etc.) in a box (bag, drawer, deck, etc.) with and without replacement is a common exercise in probability.

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Picking Probabilities -

Tag(s) : Combinatorics, Mathematics

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Picking Probabilities

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Probabilities for a Draw without Replacement

Example : Probability to pick a set of n=10 marbles with k=3 red ones (so 7 are not red) in a bag containing an initial total of N=100 marbles with m=20 red ones.





➠ Random Selection

Probabilities for multiple Draws

Example: Calculation of the probability of having drawn the number '23' after 200 drawings of a 50-face dice.





Probabilities for a Draw with Replacement

Example : Probability to pick at least once each card from a deck of N=50 cards after n=200 drawings.



➠ Random Selection

Tool to make probabilities on picking objects. Calculation of probabilities of drawing objects (balls, beads, cards, etc.) in a box (bag, drawer, deck, etc.) with and without replacement is a common exercise in probability.

Answers to Questions

How to compute a probability of picking without replacement?

Consider a set of \( N \) objects among which \( m \) are different (distinguishable). The probability of drawing a total of \( n \) objects and that among these \( n \) objects there are \( k \) objects that are part of the \( m \) different ones, is given by a hypergeometric distribution: $$ p(X=k)=\frac{C_{m}^kC_{N-m}^{n-k}}{C_N^n} = \frac{ \binom{m}{k} \binom{N-m}{n-k} }{ \binom{N}{n} } $$

C represents the combination operator.

Example: Probability to draw \( k=5 \) red card among the \( m=26 \) red cards in a deck of \( N=52 \) cards by drawing \( n=5 \) cards.

Example: Probability to draw all \( k=3 \) black ball in a bowl with \( N=25 \) balls among which \( m=3 \) are black, by picking \( n=3 \) balls.

How to compute a probability of picking with replacement?

The probability of having picked all \( N \) objects after \( n \) draws is given by the formula $$ \sum_{i=0}^N (-1)^{N-i}{\binom{N}{i}}\left(\frac{i}{N}\right)^n $$

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