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Run the experiment of tossing a fair coin 3 times 1000 times updating after every 100 tosses. To do this link here, and when the page opens click the red die in front of number 4. Set the number of coins at 3. After running the experiment 1000 times, what can you say about the theoretical (in blue) and actual (in red) probability distributions of the number of heads? In all examples of discrete random variables, the probabilities in the probability distribution table give the 'long-term' proportion of times that the random variable assumes each possible value. For examples 3 and 4, you can use this link to a simulation of the situation. When the page opens click the red die in front of number 4 to open the simulation. When the simulation opens, set N to 20, set R to 10, the number of red balls, and set n, the sample size to 2. Select with or without replacement as appropriate for the example. The blue graph and the text below it will show probabilities for each number of red balls. Consider again the count of heads in 3 tosses of a fair coin. If this experiment is repeated, say 10 times, and the number of heads in each series of 3 tosses is counted, you will have a set of numbers like 0,1,3,1,2,2,1,1,3,0. The average of these numbers is the average value for the random variable, the number of heads in 3 tosses of a fair coin. To see what happens in a larger number of runs of the experiment, again link here, and when the page opens click the red die in front of number 4. Set the number of coins at 3 and run the experiment 100 times. What is the average number of heads per 3 tosses? Now reset and run the experiment 1000 times. What is the average number of heads per 3 tosses? The long-term average number of heads is called the expected value of the random variable, the number of heads in 3 tosses of a fair coin. This expected value can be found for most random variables. Think of expected value as the average value of a random variable. There is an easier way to find the expected value of this (or any) discrete random variable. If the experiment of tossing the coin 3 times is repeated for a large number, N, times, the experiment will end in 0 heads n0 times, in 1 head n1 times, in 2 heads n2 times, and in 3 heads n3 times. The total number of heads is 0 n0 + 1 n1 + 2 n2 + 3 n3, and the average number of heads per run of the experiment is (0 n0 + 1 n1 + 2 n2 + 3 n3)/N = 0 (n0/N) + 1 (n1/N) + 2 (n2/N) + 3 (n3/N) For large N, (n0/N) ~ P[0 Heads], (n1/N) ~ P[1 Head], (n2/N) ~ P[2 Heads], (n3/N) ~ P[3 Heads], so the average number of heads per run of the experiment is 0 P[0 Heads] + 1P[1 Head] + 2 P[2 Heads] + 3 P[3 Heads] This is called the Expected Value or Mean and is denoted, for a general random variable X, by E[X]. It can be computed by Using this formula on the random variable T, the total number of heads in 3 tosses of a fair coin, you get µ=E[T] = 0 (1/8) + 1 (3/8) + 2 (3/8) + 3 (1/8) = 12/8 = 3/2 = 1.5. This can be interpreted as the average number of heads per sequence of 3 tosses if the experiment is repeated a large number of times. Just as you are able to find the average value for a random variable, so you can also find the standard deviation of the random variable. In the case of a random variable, the standard deviation is given by For random variable T, the total number of heads in 3 tosses of a fair coin, the standard deviation computed by the rightmost formula is SD[T] = Square Root of [02 (1/8) + 12 (3/8) + 22 (3/8) + 32 (1/8) - (3/2)2] = Square Root of [3/4] = 31/2/2
Tossing a Coin:
We say the probability of the coin landing H is ½
Throwing a Die:
We say the probability of a four is 1/6 (one of the six faces is a four) Note that a die has 6 sides but here we look at only two cases: "four: yes" or "four: no" Toss a fair coin three times ... what is the chance of getting exactly two Heads? Using H for heads and T for Tails we may get any of these 8 outcomes:
Which outcomes do we want?"Two Heads" could be in any order: "HHT", "THH" and "HTH" all have two Heads (and one Tail). So 3 of the outcomes produce "Two Heads". What is the probability of each outcome?Each outcome is equally likely, and there are 8 of them, so each outcome has a probability of 1/8 So the probability of event "Two Heads" is:
So the chance of getting Two Heads is 3/8
We used special words:
3 Heads, 2 Heads, 1 Head, NoneThe calculations are (P means "Probability of"):
We can write this in terms of a Random Variable "X" = "The number of Heads from 3 tosses of a coin":
And this is what it looks like as a graph: It is symmetrical! Making a FormulaNow imagine we want the chances of 5 heads in 9 tosses: to list all 512 outcomes will take a long time! So let's make a formula. In our previous example, how can we get the values 1, 3, 3 and 1 ? Well, they are actually in Pascal’s Triangle ! Can we make them using a formula? Sure we can, and here it is: The formula may look scary but is easy to use. We only need two numbers:
The "!" means "factorial", for example 4! = 1×2×3×4 = 24 Note: it is often called "n choose k" and you can learn more here. Let's try it:
We have n=3 and k=2:
n!k!(n-k)! =3!2!(3-2)! =3×2×12×1 × 1 = 3 So there are 3 outcomes that have "2 Heads" (We knew that already, but we now have a formula for it.) Let's use it for a harder question:
We have n=9 and k=5:
n!k!(n-k)! =9!5!(9-5)! =9×8×7×6×5×4×3×2×15×4×3×2×1 × 4×3×2×1 =126 So 126 of the outcomes will have 5 heads And for 9 tosses there are a total of 29 = 512 outcomes, so we get the probability:
So: P(X=5) = 126512 = 0.24609375 About a 25% chance. (Easier than listing them all.) Bias!So far the chances of success or failure have been equally likely. But what if the coins are biased (land more on one side than another) or choices are not 50/50.
This is just like the heads and tails example, but with 70/30 instead of 50/50. Let's draw a tree diagram: The "Two Chicken" cases are highlighted. The probabilities for "two chickens" all work out to be 0.147, because we are multiplying two 0.7s and one 0.3 in each case. In other words 0.147 = 0.7 × 0.7 × 0.3 Or, using exponents: = 0.72 × 0.31 The 0.7 is the probability of each choice we want, call it p The 2 is the number of choices we want, call it k And we have (so far): = pk × 0.31 The 0.3 is the probability of the opposite choice, so it is: 1−p The 1 is the number of opposite choices, so it is: n−k Which gives us: = pk(1-p)(n-k) Where
So we get:
pk(1-p)(n-k) =0.72(1-0.7)(3-2) =0.72(0.3)(1) =0.7 × 0.7 × 0.3 =0.147 which is what we got before, but now using a formula Now we know the probability of each outcome is 0.147 But we need to include that there are three such ways it can happen: (chicken, chicken, other) or (chicken, other, chicken) or (other, chicken, chicken)
The total number of "two chicken" outcomes is:
n!k!(n-k)! =3!2!(3-2)! =3×2×12×1 × 1 =3 And we get:
So the probability of event "2 people out of 3 choose chicken" = 0.441 OK. That was a lot of work for something we knew already, but now we have a formula we can use for harder questions.
So we have: And we get:
pk(1-p)(n-k) =0.77(1-0.7)(10-7) =0.77(0.3)(3) =0.0022235661 That is the probability of each outcome. And the total number of those outcomes is:
n!k!(n-k)! =10!7!(10-7)! =10×9×8×7×6×5×4×3×2×17×6×5×4×3×2×1 × 3×2×1 =10×9×83×2×1 =120 And we get:
So the probability of 7 out of 10 choosing chicken is only about 27% Moral of the story: even though the long-run average is 70%, don't expect 7 out of the next 10. Putting it TogetherNow we know how to calculate how many: n!k!(n-k)! And the probability of each: pk(1-p)(n-k) When multiplied together we get:
Probability of k out of n ways: P(k out of n) = n!k!(n-k)! pk(1-p)(n-k) The General Binomial Probability Formula Important Notes:
QuincunxHave a play with the Quincunx (then read Quincunx Explained) to see the Binomial Distribution in action. A fair die is thrown four times. Calculate the probabilities of getting:
In this case n=4, p = P(Two) = 1/6 X is the Random Variable ‘Number of Twos from four throws’. Substitute x = 0 to 4 into the formula: P(k out of n) = n!k!(n-k)! pk(1-p)(n-k) Like this (to 4 decimal places):
Summary: "for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)" This time the graph is not symmetrical: It is not symmetrical! It is skewed because p is not 0.5 Sports BikesYour company makes sports bikes. 90% pass final inspection (and 10% fail and need to be fixed). What is the expected Mean and Variance of the 4 next inspections? First, let's calculate all probabilities. X is the Random Variable "Number of passes from four inspections". Substitute x = 0 to 4 into the formula: P(k out of n) = n!k!(n-k)! pk(1-p)(n-k) Like this:
Summary: "for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection." Mean, Variance and Standard DeviationLet's calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections. There are (relatively) simple formulas for them. They are a little hard to prove, but they do work! The mean, or "expected value", is: μ = np
For the sports bikes: μ = 4 × 0.9 = 3.6 So we can expect 3.6 bikes (out of 4) to pass the inspection. The formula for Variance is: Variance: σ2 = np(1-p) And Standard Deviation is the square root of variance: σ = √(np(1-p))
For the sports bikes: Variance: σ2 = 4 × 0.9 × 0.1 = 0.36 Standard Deviation is: σ = √(0.36) = 0.6
Note: we could also calculate them manually, by making a table like this:
The mean is the Sum of (X × P(X)): μ = 3.6 The variance is the Sum of (X2 × P(X)) minus Mean2: Variance: σ2 = 13.32 − 3.62 = 0.36 Standard Deviation is:σ = √(0.36) = 0.6 And we got the same results as before (yay!)Summary
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