(27S) Probability Formulas

[Dave Britten]

Note: These should also work on the 19B and palmtops (95LX/100LX/200LX). The 17B lacks the COMB function needed for several of these, but it can make use of Balls & Urns, and Birthday Paradox.

All example results are displayed as in FIX 6.

Binomial Probability Distribution

Code:

0*(N+P1+MIN+MAX)+Σ(X:MIN:MAX:1:COMB(N:X)*P1^X*(1-P1)^(N-X))=ΣP

Calculates the total probability of achieving a number of successes within a desired range from a fixed number of independent trials with known probability of success.

Variables:

N - Number of trials
P1 - Probability of success for a single trial
MIN - Minimum desired number of successes
MAX - Maximum desired number of successes

Enter the above four variables, and solve for ΣP. MIN and MAX may be set equal to each other to calculate the probability of a specific number of successes.

Example:

I roll a fair, six-sided die 10 times. What is the percent chance that I get a 6 at least 4 times?

10 {N}
1÷6 {P1}
4 {MIN}
10 {MAX}
{ΣP} 0.069728 (i.e. 6.9728%)

Negative Binomial Probability Distribution

Code:

Σ(X:IF(NEED>MIN:NEED:MIN):MAX:1:COMB(X-1:NEED-1)*P1^NEED*(1-P1)^(X-NEED))=ΣP

Calculates the total probability of achieving a set number of successes in an indefinite number of independent trials, continuing until the desired number of successes has been reached.

Variables:

NEED - Target number of successes
MIN - Minimum number of trials
MAX - Maximum number of trials
P1 - Probability of success for a single trial

Enter the above four variables, then solve for ΣP. MIN may be set equal to NEED if you don't want a lower bound on the number of trials, i.e. compute the probability of any number of trials less than or equal to MAX. MIN and MAX may be set equal to calculate the odds of finishing at a specific number of trials.

Example: I will keep rolling a die until I've gotten a 6 four times. What is the percent chance that I will be finished within 20 rolls?

4 {NEED}
4 {MIN}
20 {MAX}
1÷6 {P1}
{ΣP} 0.433454 (i.e. 43.3454%)

Hypergeometric Probability Distribution

Code:

0*(POP+GOOD+SAM+MIN+MAX)+Σ(X:MIN:MAX:1:COMB(POP-GOOD:SAM-X)*COMB(GOOD:X)÷COMB(POP:SAM))=ΣP

In repeated trials of sampling without replacement what are the odds of achieving a target number of successes?

Variables:

POP - Population size
GOOD - Number of items among the population considered to be a success
SAM - Number of trials/samples
MIN - Minimum number of successes
MAX - Maximum number of successes

Enter the above five variables, and solve for ΣP. As in the binomial probability distribution formula, MIN and MAX may be set equal to test a specific number of successes rather than a range.

Example:

From a shuffled deck of 52 standard playing cards, I will draw 6 cards. What is the percent chance that at least 2 of them are face cards?

52 {POP}
12 {GOOD} (Three face cards in each of four suits.)
6 {SAM}
2 {MIN}
6 {MAX}
{ΣP} 0.423609 (i.e. 42.3609%)

Balls & Urns (Also sometimes called Stars & Bars)

Code:

COMB(U+B-(1*ZER?):B-1+ZER?)=C

How many ways are there to sort a given number of indistinct items into a given number of distinct categories?

Variables:

U - Number of urns (categories)
B - Number of balls (items)
ZER? - Allow categories with zero items? (1 for yes, 0 for no)

Enter the above three variables and solve for C.

Example:

How many different combinations can be rolled with 5 six-sided dice? Note that 1,1,1,2,2 is equivalent to 2,1,2,1,1, and that some combinations will be more likely to occur than others.

6 {U}
5 {B}
1 {ZER?}
{C} 330.000000

Birthday Paradox

Code:

0*L(T:1)*Σ(C:1:N-1:1:L(T:(P-C)÷P*G(T)))+1-G(T)=PCOL

When sampling with replacement from a fixed pool of items, what are the odds that an item will be taken more than once? (e.g. What are the odds that at least two people in a room of 25 will share the same birthday?)

Variables:

N - Number of samples
P - Population size (i.e. number of items to choose from)

Enter the above two variables, and solve for PCOL (probability of at least one collision).

Example:

Assuming a uniform distribution of birthdays among the population, and ignoring Feb. 29, what is the percent chance that, in a room of 26 people, at least two people will share the same birthday?

26 {N}
365 {P}
{PCOL} 0.598241 (i.e. 59.8241%)