How does the binomial probability mass function calculate the probability of exactly k successes in n independent trials?

Conditions

• The random variable X follows a binomial distribution with parameters n and p.
• k is an integer satisfying $0 \le k \le n$.
• Trials are mutually independent with a constant success probability p.

Reasoning, step by step

1. Calculate the probability of a single fixed sequence containing k successes and n-k failures, which is $p^k(1-p)^{n-k}$.
2. Determine the number of different ways to choose the positions of the k successes among the n trials, represented by the binomial coefficient $\binom{n}{k}$.
3. Multiply the single-sequence probability by the number of arrangements to get the total probability of exactly k successes.

Example

For an 82-game season with a 70% win probability, the probability of exactly 60 wins is expressed as $Pr(X=60)=\binom{82}{60}(0.7)^{60}(0.3)^{22}$. The video explains: 'The factor $p^k(1-p)^{n-k}$ gives the probability of one fixed win-and-loss sequence... The factor $\binom{n}{k}$ counts the ways to choose, from 82 games, the 60 winning positions.'

Common misconceptions

• Thinking that $p^k(1-p)^{n-k}$ alone represents the total probability of k successes, ignoring the different possible sequences.
• Confusing the binomial coefficient with the probability of a single sequence.

Watch the explanation

Connected concepts

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• For an 82-game regular season, how does the binomial model express the probability of exactly 60 wins?