: Different strategies sampling 2 out of numbers {1,2,3}: Sampling WITHOUT replacement, Order IS important Number of Samples = 6 (1,2) (1,3) (2,1) (2,3) (3,1) (3,2)
: Different strategies sampling 2 out of numbers {1,2,3}: Sampling WITHOUT replacement, Order IS NOT important Number of Samples = 3 (1,2) (1,3) (2,3) Usually, sample without replacement, thereby ensuring the sample is comprised of different individuals!
Permutations When sampling from the population WITHOUT replacement of the subjects and the order of sampled individuals IS important, the number of different arrangements, permutations of n subjects from a population of size N: where N! = [N factorial] =1*2*3* *(N-1)*N 0! = 1
: Permutations Locker combination (order is important) comprised of 3 distinct digits. Compute the number of different locker combination. N = n = P(N_n) = N!/(N-n)! =
: Permutations Locker combination (order is important) comprised of 3 distinct digits. Compute the number of different locker combination. N = 10 n = 3 P(N _ n) = N!/(N-n)! = 10!/(10-3)! = = (1*2* *7*8*9*10)/ (1*2* *7) = 720
Permutations Simple Properties General Formula
Combinations When sampling from the population WITHOUT replacement of the subjects and the order of sampled individuals IS NOT important, the number of different samples, combinations of n subjects from a population of size N:
Consider sampling 3 students from a class of 10 people. Compute the number of different samples. In this situation there will be no replacement, and the order of students in a sample is not important. N = n = C(N choose n) = N!/n!(N-n)! =
Consider sampling 3 students from a class of 10 people. Compute the number of different samples. N =10 n = 3 C(N choose n) = N!/n!(N-n)! =10!/(3!7!) = =8*9*10/(2*3) = 120
Combinations Simple Properties General Formula
You have N= close friends that you would like to invite to the movie. In how many ways can you invite 2 of them? Is order important? What do you have to compute: Permutation or Combination?
You have N= close friends that you would like to invite to the movie. In how many ways can you invite 2 of them? Order is Not important Compute Combination: C(N_2) = N*(N-1)/2
Binomial Random Variables There are n trials (performances of the binomial experiment), where n is determined, not random. Each trial results in only 1 outcome: success or failure. The probability of a success on each trial is constant, denoted p, with 0 p 1. The probability of a failure q=1-p, with 0 q 1. The trials are independent.
A coin is tossed 3 times (n=3). Each time there are two possible outcomes: tail (success) and head (failure), with constant probabilities p and q (q=1-p). Define random variable X as the number of tails in 3 independent trials. X is a discrete random variable that takes the values 0, 1, 2, 3. X ~ Binomial(n=3, p =0.5)
: Are Conditions satisfied for Binomial Model? Observe the gender of the next 50 children born at a local hospital (X= number of girls.) Yes No A ten-question quiz has 5 True-False questions and 5 multiple-choice questions, each with 4 possible choices. A student randomly picks an answer for event question. X= number of correct answers. Yes No
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X 3 2 2 2 1 1 1 0 P(X=3) =?
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X 3 2 2 2 1 1 1 0 P(X=3) = p*p*p = p^3 P(X=0) = q^3
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X 3 2 2 2 1 1 1 0 P(X=2) =?
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X 3 2 2 2 1 1 1 0 P(X=2) = p*p*q + p*q*p + q*p*p = 3p^2*q Recall C(3_2) = 3!/(2!(3-2)!) = 3
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X 3 2 2 2 1 1 1 0 P(X=1) =?
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X 3 2 2 2 1 1 1 0 P(X=1) = p*q*q + q*p*q + q*q*p = 3p*q^2 Recall C(3_1) = 3!/(1!(3-1)!) = 3
X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p X=k 0 1 2 3 P(X=k) q^3 3pq^2 3p^2q p^3
Binomial Distribution X~ Binomial (n, p) Probability of exactly k successes in n trials: where Mean (Expected value) of X is Standard Deviation of X is
A restaurant manager has calculated, based on years of experience, that about 20% of the patrons order the chef s steak special each Friday evening. Consider taking a random sample of patrons and let X be the number of sampled patrons who will order the chef s steak special on a Friday evening.
What is the exact distribution of X for a random sample of 10 patrons? (Provide all features).
What is the exact distribution of X for a random sample of 10 patrons? (Provide all features). X has a Binomial distribution with sample size n = 10 and probability of success p = 0.2.
What is the probability that exactly one patron in the sample of 10 will order the chef s steak special on a Friday evening? P(X=1) =?
What is the probability that exactly one patron in the sample of 10 will order the chef s steak special on a Friday evening? n=10, k=1, p=0.2 C(n_k) = n!/(k!(n-k)!) = 10!/(1!(10-1)!) = 10 P(X=1) =10*(0.2)^1*(1-0.2)^9 = 0.2684
What is the probability that at least one patron in the sample of 10 will order the chef s steak special on a Friday evening? P(X>=1) =?
P( X Example What is the probability that at least one patron in the sample of 10 will order the chef s steak special on a Friday evening? P(X=>1) = P(X=1) + P(X=2) + +P(X=10) Hint: Key here is to use the complement rule. P(at least one) = 1 P(none). 1) 1 P( X 0) 1 10 0 (0.2) 0 (1 0.2) 10 1 0.8 10 0.89
How many patrons in the sample of 10 you expect to order the chef s steak special on a Friday evening? For Binomial distribution E(X)=μ=np= 10*0.2=2