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# Binomial distribution

Sal introduces the binomial distribution with an example.

## Want to join the conversation?

• I'm confused with these factorials (!) In the problem around how does 5!/4!=5?
• ``5! = 5•4•3•2•14! = 4•3•2•15!   5•4•3•2•1── = ─────────4!     4•3•2•15!   ── = 54!``
• I believe Sal's approach of using 5Cx/2^5 where x is the exact number of heads only works because we have assumed a fair coin here. How would we solve this problem if, say the probability of heads on our coin was 60%? I think we would have to use something involving Bernoulli trials.
• Indeed would would use Bernoulli like trials! In particar, we'd raise 0.6 to the Xth power, and (1-0.6) to the (n-x)th power. Then we'd multiply those together and with the number of combinations. So we'd say that p=0.6, and the formula would be:

`n nCk x * p^x * (1-p)^(n-x)`
• Why are probabilitiy distributions so often simetrical?
• In this case, the graph is symmetrical because in one trial, the probability of success equals the probability of failure which equals 50%. I.e. you are equally likely to get a head or a tail when you flip a fair coin.
Say the coin was a trick coin, that gave you heads 2/3 of the time, and tails 1/3 of the time. The graph in this case would not be symmetrical.

I made a program to help with visualizing this. Try it here,
Change the probability of success from 0.5 to 0.67 (equiv to 2/3) to see what happens
• If 15 dates are selected at random, what is probability of getting two Sundays?
• depends on the range of dates they are selected from, if they are only from a range of one year there will not be exactly 1 in 7 days that are Sundays
(1 vote)
• I have a very silly question. Ya I know it's would be very silly to ask but I can't resist asking this! How can you know that for getting a consecutive 2 or 3 or 4 heads in a row, you will multiply the probability of each event.
I know that sum them up would be nonsense cause for example with 3 consecutive heads: 0.5+0.5+0.5 = 1.5 > 1.
Still I want to intuitively feel why multiplication is the best here?? Tks a lot!
• Have you looked at the tree idea before? Say the tree has a stem of 1 which we split three ways so every one of three branches has 1/3 "thickness". Now if you grow three branches out of that one, they add up to 1 * "thickness" of the parent branch. But that one isn't 1, it's 1/3 of the stem thickness. The daughter branches are each 1/3 "thickness" of the parent for example. But that must mean they are 1/3 * 1/3 * stem which I arbitrarily chose to be 1. Does that image make a little bit more sense? It's not limited to binomial distribution problems as I tried to indicate with the three branches instead of 2 (binary).
• If we have to flip a coin 10 times, what is the probability of getting tails 7 times?
• There are 10C7 = 10!∕(7! ∙ 3!) = 120 outcomes that contain 7 Tails and 3 Heads.

There are 2^10 = 1,024 possible outcomes in total.

So, the probability of flipping 7 Tails and 3 Heads is
120∕1,024 = 15∕128
• Does addition of all the values give the total probability?
• It depends on what you mean by total probability. The probability of having exactly 2 heads is 5/32, which is 15.625% or 0.15625
To arrive at that probability, we didn't do any addition.

If you mean using addition so we can find the total possible number of ways that the coins can be flipped, it is true that we can add the number of ways that there can be zero head, plus the number of ways there can be one head, plus the number of ways there can be 2 heads, plus the number of ways there can be 3 heads, plus the number of ways there can be 4 heads, plus the number of ways there can be 5 heads.
1 + 5 + 10 + 10 + 5 + 1 = 32 different outcomes when you flip a coin 5 times
If you add the fractions, you get
1/32 + 5/32 + 10/32 + 10/32 + 5/32 + 1/32 = 32/32 = 1/1 = 1
The answer of one doesn't tell you much about the coin flip outcomes, unless you are checking that the probability of zero heads plus the probability of one head plus the probability of two heads plus the probability of three heads plus the probability of four heads plus the probability of five heads will add up to 100 percent of the total outcomes.
In other words, the probability that you will get either 0, 1, 2, 3, 4, or 5 heads is 1 (which is 100%)
• I came across a question which asked me what is THE EXPANSION OF THE BIONOMIAL in algebra. What does that mean?
• The expansion (multiplying out) of (a+b)^n is like the distribution for flipping a coin n times. For the ith term, the coefficient is the same - nCi. Instead of i heads' and n-i tails', you have (a^i) * (b^(n-i)).

e.g. (a+b)^4 = a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4
(1 vote)
• To make sure I get it right "6_C_3" is the notation of 6! / [ 3! (6-3)! ], whis is interpreted as out of 6 different combination, you are interested of the of the probability of getting 3 you are intereste in. Is that correct? But we decide to sketch the tree, that is branching off in two different directions or even more? I am a little confused. For example if we had an imaginary coin with three sides, how would be the notation?
• You are right about the combination notation.
You can say, "How many possible combinations of 6 books, taken 3 at a time, exist?"
If we sketch the tree, it works pretty well for the coin-flip style questions. However, with a three-sided coin, or something similar, we would just have to treat two of the flip results as a "failure" in our "experiment," and the remaining result as a "success." That would enable us to treat the distribution as a binomial distribution.