3.2 Independent and Mutually Exclusive Events

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. The probability of picking the three of diamonds is called a conditional probability because it is conditioned on what was picked first. This is true also of the probability of picking the J of spades. The probability of picking the J of spades is actually conditioned on both the previous picks.

Flip two fair coins. (This is an experiment.)

The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.

• Let A = the event of getting at most one tail. (At most one tail means zero or one tail.) Then A can be written as {HH, HT, TH}. The outcome HH shows zero tails. HT and TH each show one tail.
• Let B = the event of getting all tails. B can be written as {TT}. B is the complement of A, so B = A′. Also, P(A) + P(B) = P(A) + P(A′) = 1.
• The probabilities for A and for B are P(A) = $\frac{3}{4}$ and P(B) = $\frac{1}{4}$.
• Let C = the event of getting all heads. C = {HH}. Since B = {TT}, $P(B\cap C)=0$. B and C are mutually exclusive. (B and C have no members in common because you cannot have all tails and all heads at the same time.)
• Let D = event of getting more than one tail. D = {TT}. P(D) = $\frac{1}{4}$
• Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = {HT, HH}. P(E) = $\frac{2}{4}$
• Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = {HT, TH, TT}. P(F) = $\frac{3}{4}$

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) = $\frac{3}{6}$ and P(B) = $\frac{3}{6}$.
• 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}. $P(C\cap D)=0$ 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}.

• Find $P\left(C\right|A)$. This is a conditional probability. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find $P\left(C\right|A)$, find the probability of C using the sample space A. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, $P\left(C\right|A)=\frac{2}{3}$.

Let event G = taking a math class. Let event H = taking a science class. Then, G $\cap$ H = taking a math class and a science class. Suppose $P\left(G\right)=0.6,P\left(H\right)=0.5$, and $P(G\cap H)=0.3.$ Are G and H independent?

If G and H are independent, then you must show ONE of the following:

• $P\left(G\right|H)=P(G)$
• $P\left(H\right|G)=P(H)$
• $P(G\cap H)=P\left(G\right)P\left(H\right)$

Since G and H are independent, knowing that a person is taking a science class does not change the chance that they are taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance they are taking math. For practice, show that $P\left(H\right|G)=P(H)$ to show that G and H are independent events.

In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.

The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes.

• $P\left(R\right)=\frac{3}{8}.P\left(B\right)=\frac{5}{8}.P(R\cap B)=0$. (You cannot draw one card that is both red and blue.)
• $P\left(E\right)=\frac{3}{8}$. (There are three even-numbered cards, R2, B2, and B4.)
• $P(E|B)=\frac{2}{5}$. (There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 and B4.)
• $P(B|E)=\frac{2}{3}$. (There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, to are blue; B2 and B4.)
• The events R and B are mutually exclusive because $P(R\cap B)=0$.
• Let G = card with a number greater than 3. G = {B4, B5}. $P\left(G\right)=\frac{2}{8}$. Let H = blue card numbered between one and four, inclusive. H = {B1, B2, B3, B4}. $P(G|H)=\frac{1}{4}$. (The only card in H that has a number greater than three is B4.) Since $\frac{2}{8}$ = $\frac{1}{4}$, $P\left(G\right)=P\left(G\right|H)$, which means that G and H are independent.

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:
• $P\left(F\right)\mathrm{=}0.60;P\left(L\right)\mathrm{=}0.50$
• $P(F\cap L)=0.45$
• $P\left(L\right|F)=0.75$

Check whether $P(F\cap L)=P\left(F\right)P\left(L\right)$. We are given that $P(F\cap L)=0.45$, but $P\left(F\right)P\left(L\right)=\left(0.60\right)\left(0.50\right)=0.30$. The events of being female and having long hair are not independent because $P(F\cap L)$ does not equal $P\left(F\right)P\left(L\right)$.

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

Interpretation of Results: The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.