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Bernoulli Experiments, Binomial Distribution
If a person randomly guesses the answers to 10 multiple choice questions, we can ask questions like I (^) what is the probability that they get none right? I (^) what is the probability that they get all ten right? I (^) what is the probability that they get at least three right? I (^) how many do they get right on average?
These and similar scenarios lead to Bernoulli Experiments and the Binomial Distribution. A Bernoulli Experiment involves repeated (in this case 10) independent trials of an experiment with 2 outcomes usually called “success” and “failure” (in this case getting a question right/wrong).
Bernoulli Experiment with n Trials
Here are the rules for a Bernoulli experiment.
- The experiment is repeated a fixed number of times (n times).
- Each trial has only two possible outcomes, “success” and “failure”. The possible outcomes are exactly the same for each trial.
- The probability of success remains the same for each trial. We use p for the probability of success (on each trial) and q = 1 − p for the probability of failure.
- The trials are independent (the outcome of previous trials has no influence on the outcome of the next trial).
- We are interested in the random variable X where X = the number of successes. Note the possible values of X are 0, 1 , 2 , 3 ,... , n.
Examples
A basketball player takes four independent free throws with a probability of 0.7 of getting a basket on each shot. The number of baskets made is recorded. Here each free throw is a trial and trials are assumed to be independent. Each trial has two outcomes basket (success) or no basket (failure). The probability of success is p = 0.7 and the probability of failure is q = 1 − p = 0.3. We are interested in the variable X which counts the number of successes in 4 trials. This is an example of a Bernoulli experiment with 4 trials.
Examples
A bag contains 6 red marbles and 4 blue marbles. Five marbles are drawn from the bag without replacement and the number of red marbles is observed. We might let a trial here consist of drawing a marble from the bag and let success be getting a red. However, this is not a Bernoulli experiment since the trials are not independent (the mix of reds and blues changes on each trial since we do not replace the marble) and the probability of success and failure vary from trial to trial.
Examples
1000 people are chosen at random from among likely Democrat primary voters in Indiana, and asked “do you plan to vote for Bernie Sanders in the upcoming primary on May 3?”. Their answers are recorded. Assuming people really are randomly selected, this is an independent repetition of the same trial 1000 times. If we declare voting for Bernie “success”, then p = the proportion of likely Democrat primary voters in Indiana who plan to vote for Bernie. If X = the number of successes, then a observation of X can be used to estimate (unknown) p. For example, if X = 442 in a particular running of this poll, we would concluded that p is likely around 44%.
Probability distribution of X
Our next goal is to calculate the probability distribution for the random variable X, where X counts the number of successes in a Bernoulli experiment with n trials. We will start with a small example for which a tree diagram can be drawn (we have already looked at a specific case of this example when we studied tree diagrams).
Example: A basketball player takes 4 independent free throws with a probability of 0.7 of getting a basket on each shot. Let X = the number of baskets he gets. Notice that this is indeed a Bernoulli experiment with n = 4 and p = 0.7.
Probability distribution of X
In general we have the following:
If X is the number of successes in a Bernoulli experiment with n independent trials, where the probability of success is p in each trial (and the probability of failure is then q = 1 − p), then
P(X = k) = C(n, k)pkqn−k^ =
n k
pkqn−k
for k = 0, 1 , 2 , · · · , n.
Probability distribution of X
We can see why this is true if we visualize a tree diagram for the n independent trials. The number of paths with exactly k success (out of n trials) is C(n, k) and the probability of every such path equals pkqn−k. The event that X = k can result from any one of these outcomes (paths), hence the P(X = k) is the sum of the probabilities of all paths with exactly k successes which is C(n, k)pkqn−k.
Example
X P(X)
0 C(4, 0)(0.7)^0 (0.3)^4 0. 0081
1 C(4, 1)(0.7)^1 (0.3)^3 0. 0756
2 C(4, 2)(0.7)^2 (0.3)^2 0. 2646
3 C(4, 3)(0.7)^3 (0.3)^1 0. 4116
4 C(4, 4)(0.7)^4 (0.3)^0 0. 2401
Note 0. 0081 +
- 0756 + 0. 2646 + 0 .4116 + 0.2401 = 1
Binomial Random Variables
For a Bernoulli experiment with n trials, let X denote the number of successes in the n trials, where the probability of success in each trial is p. This distribution of random the variable X is called a binomial distribution with parameters n and p.
The expected value of X is
E(X) = np
and the standard deviation of X is
σ(X) =
npq
where q = 1 − p.
Examples
A student is given a multiple choice exam with 10 questions, each question with five possible answers. He guesses randomly for each question.
(a) What’s P (he will get exactly 6 questions correct)? n = 10, k = 6, p = 0.2, so C(10, 6)(0.2)^6 (0.8)^4 ≈ 0. 0055
(b) What is the probability he will get at least 6?
C(10, 6)(0.2)^6 (0.8)^4 + C(10, 7)(0.2)^7 (0.8)^3 + C(10, 8)(0.2)^8 (0.8)^2 + C(10, 9)(0.2)^9 (0.8)^1 + C(10, 10)(0.2)^10 (0.8)^0 ≈ 0 .0063. (c) What is the expected number of correct answers, and what’s the standard deviation? E(X) = np = 10 · 0 .2 = 2, σ(X) =
npq =
A slightly off-topic example
Assume that the Mets and the Royals are in the world series, that the Mets have a 3/5 chance of winning any given game, and that the games are independent experiments. What is the probability of a 7 game series?
Note 1: The world series is not a Bernoulli experiment! (number of games is not fixed in advance) Note 2 A seven game series will occur only when each team wins 3 of the first 6 games. A seven game series will occur whenever the Mets win exactly 3 of the first 6 games. The probability of this is C(6, 3)(0.6)^3 (0.4)^3 ≈ 0 .27. It is also true that a seven game series will occur whenever the Royals win exactly 3 of the first 6 games. The probability of this is C(6, 3)(0.4)^3 (0.6)^3 ≈ 0 .27.
Polling example
Suppose that the voting population in Utopia is 300 million and 60% of the voting population intend to vote for Melinda McNulty in the next election. We take a random sample of size 100 from the same voting population and ask each person chosen whether they will vote for Melinda McNulty in the next election or not. Let X be the number of YESes in our sample. The possible values of X (the number of successes) are 0, 1 , 2 , 3 , · · · , 100. n = 100, p = 0. 6 , q = 1 − p = 0.4. Calculate the following:
(a) What is P(X = 60)? (b) What is P(X 6 20)?
(c) What is P(X > 70)? (d) What is P(X < 50)?
(e) What is P(50 6 X 6 60)?
Polling example
P(X = 60) ≈ 0. 0812
P(X 6 20) ≈ 3. 42 × 10 −^16
P(X > 70) = 1 − P(X 6 70) ≈ 0. 0147
P(X < 50) = P(X 6 49) ≈ 0. 0167
P((50 6 X 6 60) =
P(X = 50) + P(X = 51) + · · · + P(X = 60)
OR
P(50 6 X 6 60) = P(X 6 60) − P(X 6 49) ≈
