If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Binomial variables

An introduction to a special class of random variables called binomial random variables.

## Want to join the conversation?

• why is the probability of getting heads 0.6 and not 0.5?
• Not all coins are "fair". An example of an unfair coin is one that has two heads; another might be one that is weighted to fall on heads more frequently than tails.

I believe Sal's point to introducing a coin with P(heads)=0.6 is to start viewers out with the idea of trials having a successful or failed outcome that is easy to understand (lots of videos with flipped coins) and is also in keeping with random variables' outcomes having different probabilities.
• Aren't the probability of success is constant, and being independent trials the same ?
• No.

Say you have 2 coins, and you flip them both (one flip = 1 trial), and then the Random Variable X = # heads after flipping each coin once (2 trials).
However, unlike the example in the video, you have 2 different coins, coin 1 has a 0.6 probability of heads, but coin 2 has a 0.4 probability of heads. This WOULD satisfy the requirement of the trials being independent, but not the requirement of the probability of success being the same for each trial.

Hope this helps!
• What does the condition 'finite number of independent trials' mean?
• This means that the number of trials is finite (instead of infinite), and the results of any trials do not affect the success probabilities for other trials.

Have a blessed, wonderful day!
• It is not true when he says "If I get a king that looks like that would be a success. If I don't get a king that would be a failure. So it seems to meet that right over there." because there is chances that both the cards that he picks up are kings. There aren't two possible outcomes there are three possible outcomes:
⚫ 0 kings are picked
⚫ 1 king is picked
⚫ 2 kings are picked
• I was thinking the same while watching the video. For me, the way the random variable was defined does not satisfy that condition. But I guess we can always define a success event as either getting a king or two kings, and the other two outputs as failure.
• Should Y be considered a binomial variable (even without replacement) beause of the "10% Rule" for assuming independence between trials? There is a related video where the 10% rule is explained.
• Even with the "10% Rule" for assuming independence between trials, Y cannot be considered a binomial variable without replacement because the probability of success (drawing a king) on each trial is not constant and the trials are not independent. The probability of success on the second trial depends on the outcome of the first trial due to the lack of replacement. Without replacement, the probability of drawing a king changes for each trial, violating the conditions for a binomial variable.
(1 vote)
• Doesn't `Point1` imply `Point4` ? Or am I missing out on some kind of explicit nuance?

For example, in Sal's first coin example, isn't the possibility of someone replacing coins (which will affect the constancy of the probability of each trial) a form of dependence?
• Same as what I replied to Mohamed, No.

Say you have 2 coins, and you flip them both (one flip = 1 trial), and then the Random Variable X = # heads after flipping each coin once (2 trials).
However, unlike the example in the video, you have 2 different coins, coin 1 has a 0.6 probability of heads, but coin 2 has a 0.4 probability of heads. This WOULD satisfy the requirement of the trials being independent, but not the requirement of the probability of success being the same for each trial.

Hope this helps!
• For a variable to be binomial, does each trial need to have the same probability of success, or is it enough that each trial has an independent constant probability of success? In other words, if I flip two coins in succession, the first one fair and the second one unfair, would that qualify as a binomial variable or not?
• The probability of success must be the same for each trial.
• What's the intuition behind it needing to have a fixed number of trials?
• The requirement of a fixed number of trials ensures that the binomial variable has a well-defined number of events over which the probability distribution can be calculated. This condition allows for the application of specific formulas and statistical methods tailored to binomial variables, providing a clear framework for analysis.
(1 vote)
• Why not introduce bernoulli distribution first? At , something seems wrong with the 1st point (made up of indepentant trials) P(K on 1st trial) = 4/52 and P(K on 2nd trial) = 4/52 * 3/51 + 48/52*4/51 => 4/52. It is the same.
(1 vote)
• Introducing the Bernoulli distribution before discussing binomial variables could indeed provide a helpful foundation. The Bernoulli distribution deals with the probability of success or failure in a single trial, which is a fundamental concept underlying binomial variables. Understanding the Bernoulli distribution can help clarify the conditions and characteristics of binomial variables. Regarding the calculation for variable Y, the probability of drawing a king on the second trial does depend on the outcome of the first trial, as there is no replacement. Therefore, the probability calculation for Y without replacement differs from the case with replacement.
(1 vote)
• It seems to be same as the multiplication rule of probability.
(1 vote)
• Yes, the calculation of the probability for variable Y without replacement resembles the multiplication rule of probability. In this case, the probability of drawing a king on the second trial depends on whether a king was drawn on the first trial, leading to a conditional probability calculation.
(1 vote)