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I would like some help reviewing Binomial Distribution, specifically when using the Probability Mass Function equation. I read the Wikipedia page and I'm not understanding how to use the Probability Mass Function equation.

K is the number of successes, right? N is the number of trials, right? N might be the number of dice rolled, coins flipped, tests performed, etc. But when I plug in the numbers for my variables I'm not getting results I would've expected.

Like for example I recall encountering an explanation about coin flips, if you flip a coin there's a 50/50 chance of getting heads, right? So, if you flip two coins there's a 75% chance of getting heads with at least one of the coins right? Or perhaps a better way of saying that would be that there's only a 25% chance of both coins landing heads up or both tails up, right?

Though when I input K = 1 (at least 1 heads up) and N = 2 (two coins flipped) I just get .125. I know that each individual coin flip has the same percent chance of getting a given result assuming it's a perfectly balanced coin that never lands on its edge, but wouldn't more coins or more trials or whatever naturally increase the likelihood of a given event?

Question clarification:

Sample Probability question: what is the percent chance of seeing at least one heads up result if you flip two coins at the same time?

My work: (.5^2)(1-.5)^2-1 = .25(.5)^1 = .125

My thoughts: There cannot be a 12.5% chance of seeing at least 1 heads up toss between 2 flipped coins. I know that I'm doing something wrong, but I don't know what. (I got two other responses regarding cumulative distribution functions, but I am not familiar with those).

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  • $\begingroup$ what do you meant by "when I plug in the numbers for my variables I'm not getting results I would've expected"? It will be better if you write the problem which you can't solve. $\endgroup$ – Mr.M Jul 11 '18 at 16:04
  • $\begingroup$ $\texttt{"Though when I input K = 1 (at least 1 heads up)}$ $\texttt{and N = 2 (two coins flipped) I just get .5"}$ Please show your calculation. I´m not sure what you mean. $\endgroup$ – callculus Jul 11 '18 at 16:04
  • $\begingroup$ I taught a course similar to this one recently. You are not the first person to get confused using the binomial PDF (or PMF) . // It tends to go more smoothly if you read/review all of the related examples in the text and class notes before you start with the problems. $\endgroup$ – BruceET Jul 12 '18 at 0:32
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The probability mass function you evaluated is the probability that you get exactly one head.

You may be confusing probability mass functions with distribution functions, which are also sometimes called cumulative distribution functions. These evaluate the probability that some random variable is less than or equal to $ x $.

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Your only mistake is a wrong use of the parameter $K$. When you input $1$ as $K$ in the binomial mass function, the output is the probability of getting exactly one sucess out of $N$ trials. Your desired answer is obtained if you calculate the function for each $1\leq K\leq N$ and summing. Alternatively, you can look after the cumulative probability distribution, maybe it will fit better to your intuition.

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Let $X$ be the number of Heads in two tosses of a fair coin. Then $X \sim \mathsf{Binom}(n = 2,\; p = 1/2)$ and you seek $$P(X \ge 1) = 1 - P(X = 0) = 1 - {2 \choose 0}(.5)^0(1-.5)^2 = 1 - \frac 1 4 = \frac 3 4.$$

Do you know how to evaluate ${2 \choose 0} = \frac{2!}{0! \cdot 2!} = 1?\;$ What is $0!\,?$

For the future, you might also learn how to do such computations on a statistical calculator or using computer software. Relevant computations in R statistical software, where dbinom is a binomial PDF:

sum(dbinom(1:2, 2, 1/2))  # summing probabilities of two disjoint events: P(X=1) + P(X=2)
## 0.75
1 - dbinom(0, 2, 1/2)     # use complement rule, subtract from 1
## 0.75
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