# help with Binomial distributions and random variable tables. [closed]

when it is preferred to use binomial distribution instead of forming random variable table. actually in my book these two topics are two different chapters. and they confuse me a lot. My teacher says we have can not use combinations when question talks about replacing object. combinations are only applicable to "without replacement" type questions. is it true? this could be a silly question because i'm confused with probability. any suggestion regarding how to study probability will be greatly appreciated. please help me with probability. i am an undergraduate student. thank you.

## closed as unclear what you're asking by Dominik, Pierre-Guy Plamondon, hardmath, suomynonA, ShaileshDec 22 '16 at 0:03

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• please help i know this is not a good question. but i can't understand probability. – Pushkar Soni Dec 21 '16 at 10:50
• Maybe your book has separate discussions of binomial (with replacement) and hypergeometric (without replacement) distributions. In @GrahamKemp's Answer, $X$ is binomial and $Y$ is hypergeometric. – BruceET Dec 21 '16 at 13:23
• While probability may not be a strong subject for you, surely you as an undergraduate can do more with punctuation and capitalization to make the request for help intelligible. Focus on stating what is known and what has to be derived. – hardmath Dec 21 '16 at 23:07

A binomial distribution is that of the count of successes in a sequence of identical and independent Bernoulli success/failure trials with a known amount of trials, $n$, and success rate, $p$.   You will only use a binomial distribution when your random variable can be described as such a count, otherwise you do not.
The use of combinations (by which I presume you mean the binomial coefficient, $^n\rm C_k$ or $\tbinom nk$) in probability calculations depends on what you are counting; how, when, and why.
For instance, if $X$ is the count of red balls selected in 5 draws with replacement from a bag of 4 red and 6 green balls, then the probability mass function is:$$\mathsf P(X=k)~=~\binom 5 k \frac{4^k~2^{5-k}}{10^5}\quad\bigl[k\in \{0,..,5\}\bigr]$$
... in which the coefficient count the arrangement of $k$ successes and $5-k$ failures, which is then multiplied by the probability of obtaining a particular arrangement.
Conversely, if $Y$ is the count of red balls selected in a 5 draws without replacement from a bag of 4 red and 6 green balls, then the probability mass function is:$$\mathsf P(Y=k)~=~\frac{\binom 4k\binom 6{5-k}}{\binom {10}5}\quad\bigl[k\in\{0,..,4\}\bigr]$$
... in which the coefficients count, in the numerator, the ways to select $k$ from $4$ red balls and $n-k$ from $6$ green balls versus, in the denominator, the ways to select any $5$ from all $10$ balls.