STATS 3.2

In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent? The following probabilities are given in this example: P(F) = 0.60; P(L) = 0.50 P(F AND L) = 0.45 P(L|F) = 0.75

Check whether P(L|F) equals P(L). We are given that P(L|F) = 0.75, but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.

Without replacement

When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.

To show two events are independent, you must show

only one of the above conditions.

Two events A and B are independent if the

the knowledge that one occurred does not affect the chance the other occurs

Sampling may be done

with replacement or without replacement.

If two events are NOT independent

then we say that they are dependent.

If it is not known whether A and B are mutually exclusive, assume

they are not until you can show otherwise

With replacement

If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.

Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}. Find the complement of A, A′. The complement of A, A′, is B because A and B together make up the sample space. P(A) + P(B) = P(A) + P(A′) = 1. Also, P(A) = 3636 and P(B) = 3636. Let event C = odd faces larger than two. Then C = {3, 5}. Let event D = all even faces smaller than five. Then D = {2, 4}. P(C AND D) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events. Let event E = all faces less than five. E = {1, 2, 3, 4}. Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?

No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = 1/6. To be mutually exclusive, P(C AND E) must be zero.

Two events are independent if the following are true:

P(A|B) = P(A) P(B|A) = P(B) P(A AND B) = P(A)P(B)

Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H independent? If G and H are independent, then you must show ONE of the following: P(G|H) = P(G) P(H|G) = P(H) P(G AND H) = P(G)P(H)

P(G|H) = 𝑃(𝐺 AND 𝐻)𝑃(𝐻)P(G AND H)P(H) = 0.3/0.5 = 0.6 = P(G)

Independent and mutually exclusive

do not mean the same thing

Let event C = taking an English class. Let event D = taking a speech class. Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225. Justify your answers to the following questions numerically. Are C and D independent? Are C and D mutually exclusive? What is P(D|C)?

Yes, because P(C|D) = P(C). No, because P(C AND D) is not equal to zero. P(D|C) = 𝑃(𝐶 AND 𝐷)𝑃(𝐶)P(C AND D)P(C) = 0.225/0.75 = 0.3

a. Toss one fair coin (the coin has two sides, H and T). The outcomes are ________. Count the outcomes. There are ____ outcomes. b. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes. c. Multiply the two numbers of outcomes. The answer is _______. d. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in part c. is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H1 and T6.) e. Event A = heads (H) on the coin followed by an even number (2, 4, 6) on the die.A = {_________________}. Find P(A). f. Event B = heads on the coin followed by a three on the die. B = {________}. Find P(B). g. Are A and B mutually exclusive? (Hint: What is P(A AND B)? If P(A AND B) = 0, then A and B are mutually exclusive.) h. Are A and B independent? (Hint: Is P(A AND B) = P(A)P(B)? If P(A AND B) = P(A)P(B), then A and B are independent. If not, then they are dependent).

a. H and T; 2 b. 1, 2, 3, 4, 5, 6; 6 c. 2(6) = 12 d. T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6 e. A = {H2, H4, H6}; P(A) = 3/12 f. B = {H3}; P(B) = 1/12 g. Yes, because P(A AND B) = 0 h. P(A AND B) = 0.P(A)P(B) = (3/12)(11/2). P(A AND B) does not equal P(A)P(B), so A and B are dependent.

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit.

a. Sampling with replacement:Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are {Q of spades, ten of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck. b. Sampling without replacement:Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are {K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. a. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD. b. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH, 7D, 6D, KH. Which of a. or b. did you sample with replacement and which did you sample without replacement?

a. Without replacement; b. With replacement

Flip two fair coins. Find the probabilities of the events. a. Let F = the event of getting at most one tail (zero or one tail). b. Let G = the event of getting two faces that are the same. c. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. d. Are F and G mutually exclusive? e.Let J = the event of getting all tails. Are J and H mutually exclusive?

a. Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = 3434 b. Two faces are the same if HH or TT show up. P(G) = 2424 c. A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P(H) = 2424 d. F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive. e. Getting all tails occurs when tails shows up on both coins (TT). H's outcomes are HH and HT. J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive.

A and B are mutually exclusive events if they

cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0.

If it is not known whether A and B are independent or dependent, assume they are

dependent until you can show otherwise.