In probability formula, what does n represent?

AI Thread Summary
In the probability formula PN(n) = N!/(n!(N-N)!) (p^n)(q^(N-n)), N represents the total number of trials or elements, while little n signifies the number of successful outcomes among those trials. The formula calculates the probability of achieving exactly n successes in N independent trials, where each trial has a success probability of p and a failure probability of q = 1 - p. Examples include flipping a coin, rolling a die, or answering multiple-choice questions, where n corresponds to the number of desired successful outcomes. This understanding clarifies how to apply the formula in various scenarios. The discussion highlights the practical applications of the binomial distribution in probability calculations.
stepheckert
Messages
6
Reaction score
0
In the formula,

PN(n) = N!/(n!(N-N)!) (p^n)(q^(N-n))

N represents the number of elements, but what exactly does little n represent?

Thanks for any help!
 
Physics news on Phys.org
The formula you gave,
P_N(n) = \frac{N!}{n! (N - n)!} p^n q^{N - n}
is the probability of n successes after N trials in a binomial distribution.

That is, you run N trials and for each of them the chance for "success" is p, and the chance for "failure" is q = 1 - p. The formula above gives the probability that out of the N trials, you get 0 \le n \le N "successes".

Some examples: if you flip a coin N times, and define "success" as "heads comes up" (with p = 1/2) then PN(n) is the probability of finding exactly n heads and N - n tails in N flips.

If you throw a die N times, and define "success" as "six comes up" (with p = 1/6) then PN(n) is the probability of getting exactly n sixes and N - n other values in N throws.

If a multiple-choice exam has N questions, and you randomly select one of the four answers for each of them then "success" can be "question is answered correctly" (with p = 1/4) and PN(n) is the probability of getting exactly n questions right and N - n wrong.
 
This helps so much! Thank you!
 
Back
Top