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8.5: Independent Events - Mathematics


Learning Objectives

In this section, you will:

  1. Define independent events
  2. Identify whether two events are independent or dependent

In the last section, we considered conditional probabilities. In some examples, the probability of an event changed when additional information was provided. This is not always the case. The additional information may or may not alter the probability of the event.

In Example (PageIndex{1}) we revisit the discussion at the beginning of the previous section and then contrast that with Example (PageIndex{2}).

Example (PageIndex{1})

A card is drawn from a deck. Find the following probabilities.

  1. The card is a king.
  2. The card is a king given that the card is a face card.

Solution

a. Clearly, (P)(The card is a king) = 4/52 = 1/13.

b. To find (P)(The card is a king | The card is a face card), we reason as follows:

There are 12 face cards in a deck of cards. There are 4 kings in a deck of cards.

(P)(The card is a king | The card is a face card) = 4/12 = 1/3.

The reader should observe that in the above example,

(P)(The card is a king | The card is a face card) ( eq) (P)(The card is a king)

In other words, the additional information, knowing that the card selected is a face card changed the probability of obtaining a king.

Example (PageIndex{2})

A card is drawn from a deck. Find the following probabilities.

  1. The card is a king.
  2. The card is a king given that a red card has shown.

Solution

a. To find (P)(The card is a king | A red card has shown), we reason as follows:

Since a red card has shown, there are only twenty six possibilities. Of the 26 red cards, there are two kings. Therefore,

(P)(The card is a king | A red card has shown) = 2/26 = 1/13.

The reader should observe that in the above example,

(P)(The card is a king | A red card has shown) = (P)(The card is a king)

In other words, the additional information, a red card has shown, did not affect the probability of obtaining a king.

Whenever the probability of an event (E) is not affected by the occurrence of another event (F), and vice versa, we say that the two events (E) and (F) are independent. This leads to the following definition.

Definition: Independent

Two Events (E) and (F) are independent if and only if at least one of the following two conditions is true.

  1. (mathbf{P}(mathbf{E} | mathbf{F})=mathbf{P}(mathbf{E})) or
  2. (mathbf{P}(mathbf{F} | mathbf{E})=mathbf{P}(mathbf{F}))

If the events are not independent, then they are dependent.

If one of these conditions is true, then both are true.

We can use the definition of independence to determine if two events are independent.

We can use that definition to develop another way to test whether two events are independent.

Recall the conditional probability formula:

[mathrm{P}(mathrm{E} | mathrm{F})=frac{mathrm{P}(mathrm{E} cap mathrm{F})}{mathrm{P}(mathrm{F})} onumber]

Multiplying both sides by (mathrm{P}(mathrm{F})), we get

[mathrm{P}(mathrm{E} cap mathrm{F})=mathrm{P}(mathrm{E} | mathrm{F}) mathrm{P}(mathrm{F}) onumber]

Now if the two events are independent, then by definition

[mathrm{P}(mathrm{E} | mathrm{F})=mathrm{P}(mathrm{E}) onumber]

Substituting, (P(E cap F)=P(E) P(F))

We state it formally as follows.

Test For Independence

Two events (E) and (F) are independent if and only if

[mathbf{P}(mathbf{E} cap mathbf{F})=mathbf{P}(mathbf{E}) mathbf{P}(mathbf{F}) onumber]

In the Examples (PageIndex{3}) and (PageIndex{4}), we’ll examine how to check for independence using both methods:

  • Examine the probability of intersection of events to check whether (P(E cap F)=P(E) P(F))
  • Examine conditional probabilities to check whether (P(E | F)=P(E)) or (P(F|E)=P(F))

We need to use only one of these methods. Both methods, if used properly, will always give results that are consistent with each other.

Use the method that seems easier based on the information given in the problem.

Example (PageIndex{3})

The table below shows the distribution of color-blind people by gender.

Male(M)

Female(F)

Total

Color-Blind(C)

6

1

7

Not Color-Blind(N)

46

47

93

Total

52

48

100

where (M) represents male, (F) represents female, (C) represents color-blind, and (N) represents not color-blind. Are the events color-blind and male independent?

Solution 1: According to the test for independence, (C) and (M) are independent if and only if (mathrm{P}(mathrm{C} cap mathrm{M})=mathrm{P}(mathrm{C}) mathrm{P}(mathrm{M})).

From the table: (P(C)) = 7/100, (P(M)) = 52/100 and (P(C cap M)) = 6/100

So (P(C) P(M)) = (7/100)(52/100) = .0364

which is not equal to (P(C cap M)) = 6/100 = .06

Therefore, the two events are not independent. We may say they are dependent.

Solution 2: (C) and (M) are independent if and only if (P(C|M) = P(C)).

From the total column (P(C)) = 7/100 = 0.07

From the male column (P(C|M)) = 6/52= 0.1154

Therefore (P(C|M) eq P(C)), indicating that the two events are not independent.

Example (PageIndex{4})

In a city with two airports, 100 flights were surveyed. 20 of those flights departed late.

  • 45 flights in the survey departed from airport A; 9 of those flights departed late.
  • 55 flights in the survey departed from airport B; 11 flights departed late.

Are the events "depart from airport A" and "departed late" independent?

Solution 1

Let A be the event that a flight departs from airport A, and L the event that a flight departs late. We have

(P(A cap L)) = 9/100, (P(A)) = 45/100 and (P(L)) = 20/100

In order for two events to be independent, we must have (P(A cap L) = P(A) P(L))

Since (P(A cap L)) = 9/100 = 0.09

and (P(A) P(L)) = (45/100)(20/100) = 900/10000 = 0.09

the two events "departing from airport A" and "departing late" are independent.

Solution 2

The definition of independent events states that two events are independent if (P(E|F)=P(E)).

In this problem we are given that

(P(L|A)) = 9/45= 0.2 and (P(L)) = 20/100 = 0.2

(P(L|A) = P(L)), so events "departing from airport A" and "departing late" are independent.

Example (PageIndex{5})

A coin is tossed three times, and the events (E), (F) and (G) are defined as follows:

(E): The coin shows a head on the first toss.

(F): At least two heads appear.

(G): Heads appear in two successive tosses.

Determine whether the following events are independent.

  1. (E) and (F)
  2. (F) and (G)
  3. (E) and (G)

Solution

We list the sample space, the events, their intersections and the probabilities.

egin{aligned}
&mathrm{S}={mathrm{HHH}, mathrm{HHT}, mathrm{HTH}, mathrm{HTT}, mathrm{THH}, mathrm{THT}, mathrm{TTH}, mathrm{TTT}
&egin{array}{ll}
mathrm{E}={mathrm{HHH}, mathrm{HHT}, mathrm{HTH}, mathrm{HTT}}, & mathrm{P}(mathrm{E})=4 / 8 ext { or } 1 / 2
mathrm{F}={mathrm{HHH}, mathrm{HHT}, mathrm{HTH}, mathrm{THH}}, & mathrm{P}(mathrm{F})=4 / 8 ext { or } 1 / 2
mathrm{G}={mathrm{HHT}, mathrm{THH}}, & mathrm{P}(mathrm{G})=2 / 8 ext { or } 1 / 4
mathrm{E} cap mathrm{F}={mathrm{HHH}, mathrm{HHT}, mathrm{HTH}}, & mathrm{P}(mathrm{E} cap mathrm{F})=3 / 8
mathrm{F} cap mathrm{G}={mathrm{HHT}, mathrm{THH}}, & mathrm{P}(mathrm{F} cap mathrm{G})=2 / 8 ext { or } 1 / 4
mathrm{E} cap mathrm{G}={mathrm{HHT}} & mathrm{P}(mathrm{E} cap mathrm{G})=1 / 8
end{array}
end{aligned}

a. (E) and (F) will be independent if and only if (P(E cap F) = P(E) P(F))

(P(E cap F) = 3/8) and (P(E) P(F) = 1/2 cdot 1/2 = 1/4).

Since 3/8 ≠ 1/4, we have (P(E cap F) eq P(E) P(F)).

Events (E) and (F) are not independent.

b. (F) and (G) will be independent if and only if (P(F cap G) = P(F) P(G)).

(P(F cap G) = 1/4) and (P(F) P(G) = 1/2 cdot 1/4 =1/8).

Since 3/8 ≠ 1/4, we have (P(F cap G) eq P(F) P(G)).

Events (F) and (G) are not independent.

c. (E) and (G) will be independent if (P(E cap G) = P(E) P(G))

(P(E cap G) = 1/8) and (P(E) P(G) = 1/2 cdot 1/4 =1/8)

Events (E) and (G) are independent events because (P(E cap G) = P(E) P(G))

Example (PageIndex{6})

The probability that Jaime will visit his aunt in Baltimore this year is .30, and the probability that he will go river rafting on the Colorado river is .50. If the two events are independent, what is the probability that Jaime will do both?

Solution

Let (A) be the event that Jaime will visit his aunt this year, and (R) be the event that he will go river rafting.

We are given (P(A)) = .30 and (P(R)) = .50, and we want to find (P(A cap R)).

Since we are told that the events (A) and (R) are independent,

[P(A cap R)=P(A) P(R)=(.30)(.50)=.15 onumber]

Example (PageIndex{7})

Given (P(B | A) = .4). If A and B are independent, find (P(B)).

Solution

If (A) and (B) are independent, then by definition (P(B | A) = P(B))

Therefore, (P(B) = .4)

Example (PageIndex{8})

Given (P(A) =.7), (P(B| A) = .5). Find (P(A cap B)).

Solution 1

By definition (P(B | A)=frac{P(A cap B)}{P(A)})

Substituting, we have

[.5=frac{mathrm{P}(mathrm{A} cap mathrm{B})}{.7} onumber]

Therefore, (P(A cap B) = .35)

Solution 2

Again, start with (P(B | A)=frac{P(A cap B)}{P(A)})

Multiplying both sides by (P(A)) gives

[P(A cap B)=P(B | A) P(A)=(.5)(.7)=.35 onumber]

Both solutions to Example (PageIndex{8}) are actually the same, except that in Solution 2 we delayed substituting the values into the equation until after we solved the equation for (P(A cap B)). That gives the following result:

Multiplication Rule for events that are NOT independent

If events (E) and (F) are not independent

[mathbf{P}(mathbf{E} cap mathbf{F})=mathbf{P}(mathbf{E} | mathbf{F}) mathbf{P}(mathbf{F}) quad ext { and } quad mathbf{P}(mathbf{E} cap mathbf{F})=mathbf{P}(mathbf{F} | mathbf{E}) mathbf{P}(mathbf{E}) onumber]

Example (PageIndex{9})

Given (P(A) =.5), (P(A cup B ) = .7), if (A) and (B) are independent, find (P(B)).

Solution

The addition rule states that

[mathrm{P}(mathrm{A} cup mathrm{B})=mathrm{P}(mathrm{A})+mathrm{P}(mathrm{B})-mathrm{P}(mathrm{A} cap mathrm{B}) onumber]

Since (A) and (B) are independent, (P(A cap B)=P(A) P(B))

We substitute for (P(A cap B)) in the addition formula and get

[mathrm{P}(mathrm{A} cup mathrm{B})=mathrm{P}(mathrm{A})+mathrm{P}(mathrm{B})-mathrm{P}(mathrm{A}) mathrm{P}(mathrm{B}) onumber]

By letting (P(B) = x), and substituting values, we get

[egin{array}{l}
.7=.5+x-.5 x
.7=.5+.5 x
.2=.5 x
.4=x
end{array} onumber]

Therefore, (P(B) = .4)


Learn About Independent Events With Example Problems And Interactive Exercises

Experiment 1: A dresser drawer contains one pair of socks with each of the following colors: blue, brown, red, white and black. Each pair is folded together in a matching set. You reach into the sock drawer and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the red pair of socks both times?

There are a couple of things to note about this experiment. Choosing a pairs of socks from the drawer, replacing it, and then choosing a pair again from the same drawer is a compound event. Since the first pair was replaced, choosing a red pair on the first try has no effect on the probability of choosing a red pair on the second try. Therefore, these events are independent.

Definition: Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring.

Some other examples of independent events are:

  • Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die.
  • Choosing a marble from a jar AND landing on heads after tossing a coin.
  • Choosing a 3 from a deck of cards, replacing it, AND then choosing an ace as the second card.
  • Rolling a 4 on a single 6-sided die, AND then rolling a 1 on a second roll of the die.

To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically below. Note that multiplication is represented by AND.

Multiplication Rule 1: When two events, A and B, are independent, the probability of both occurring is:

(Note: Another multiplication rule will be introduced in the next lesson.) Now we can apply this rule to find the probability for Experiment 1.

Experiment 1: A dresser drawer contains one pair of socks with each of the following colors: blue, brown, red, white and black. Each pair is folded together in a matching set. You reach into the sock drawer and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the red pair of socks both times?

P(red) = 1
5
P(red and red) = P(red) · P(red)
= 1 · 1
5 5
= 1
25

Experiment 2: A coin is tossed and a single 6-sided die is rolled. Find the probability of landing on the head side of the coin and rolling a 3 on the die.

P(head) = 1
2
P(3) = 1
6
P(head and 3) = P(head) · P(3)
= 1 · 1
2 6
= 1
12

Experiment 3: A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and then an eight?

P(jack) = 4
52
P(8) = 4
52
P(jack and 8) = P(jack) · P(8)
= 4 · 4
52 52
= 16
2704
= 1
169

Experiment 4: A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. What is the probability of choosing a green and then a yellow marble?

P(green) = 5
16
P(yellow) = 6
16
P(green and yellow) = P(green) · P(yellow)
= 5 · 6
16 16
= 30
256
= 15
128

Each of the experiments above involved two independent events that occurred in sequence. In some cases, there was replacement of the first item before choosing the second item this replacement was needed in order to make the two events independent. Multiplication Rule 1 can be extended to work for three or more independent events that occur in sequence. This is demonstrated in Experiment 5 below.

Experiment 5: A school survey found that 9 out of 10 students like pizza. If three students are chosen at random with replacement, what is the probability that all three students like pizza?

P(student 1 likes pizza) = 9
10
P(student 2 likes pizza) = 9
10
P(student 3 likes pizza) = 9
10
P(student 1 and student 2 and student 3 like pizza) = 9 · 9 · 9 = 729
10 10 10 1000

All of the experiments above involved independent events with a small population (e.g. A 6-sided die, a 2-sided coin, a deck of 52 cards). When a small number of items are selected from a large population without replacement, the probability of each event changes so slightly that the amount of change is negligible. This is illustrated in the following problem.

Problem: A nationwide survey found that 72% of people in the United States like pizza. If 3 people are selected at random, what is the probability that all three like pizza?

Solution: Let L represent the event of randomly choosing a person who likes pizza from the U.S.

P(L) · P(L) · P(L) = (0.72)(0.72)(0.72) = 0.37 = 37%

In the next lesson, we will address how to handle non-replacement in a small population.

Summary: The probability of two or more independent events occurring in sequence can be found by computing the probability of each event separately, and then multiplying the results together.

Exercises

Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.


How to Prove Markov’s Inequality and Chebyshev’s Inequality

(a) Let $X$ be a random variable that takes only non-negative values. Prove that for any $a > 0$,
[P(X geq a) leq frac.] This inequality is called Markov’s inequality.

(b) Let $X$ be a random variable with finite mean $mu$ and variance $sigma^2$. Prove that for any $a >0$,
[Pleft(|X – mu| geq a ight) leq frac.] This inequality is called Chebyshev’s inequality.


Dependent Events

But some events can be "dependent" . which means they can be affected by previous events.

Example: Drawing 2 Cards from a Deck

After taking one card from the deck there are less cards available, so the probabilities change!

Let's look at the chances of getting a King.

For the 1st card the chance of drawing a King is 4 out of 52

  • If the 1st card was a King, then the 2nd card is less likely to be a King, as only 3 of the 51 cards left are Kings.
  • If the 1st card was not a King, then the 2nd card is slightly more likely to be a King, as 4 of the 51 cards left are King.

This is because we are removing cards from the deck.

Replacement: When we put each card back after drawing it the chances don't change, as the events are independent.

Without Replacement: The chances will change, and the events are dependent.


Identifying Independent and Dependent Events

The example in the introduction demonstrated events that were clearly independent. However, it can sometimes be a challenge to identify whether events are independent or not. Consider the following example:

There are 3 green marbles and 5 blue marbles in a bag. Two marbles are drawn from the bag at random. Let G G G be the event that the first marble drawn is green. Let B B B be the event that the second marble drawn is blue. Are the events independent?

Case 1: G G G happens

When the first marble drawn is green, there are 7 7 7 marbles left in the bag, and 5 5 5 of them are blue. In this case, P ( B ) = 5 7 P(B)=dfrac<5> <7>P ( B ) = 7 5 ​ .

Case 2: G G G does not happen

When the first marble drawn is blue, there are 7 7 7 marbles left in the bag, and 4 4 4 of them are blue. In this case, P ( B ) = 4 7 P(B)=dfrac<4> <7>P ( B ) = 7 4 ​ .

The incidence of G G G affects the probability of B B B . Therefore, these events are not independent. In other words, they are dependent.

In the previous example, the first marble drawn affected which marbles were left in the bag. Whenever events happen in sequence, and the incidence of an event affects the sample space of the next event, the events will be dependent.

Events do not have to occur in sequence to be dependent. Consider this example:

When trying to determine whether events are dependent or independent, consider how the incidence of one event affects the probability of the other. If the probability is affected, then the events are dependent. If there is no effect on the probability, then the events are independent.


Example 2

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.

  1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD.
  2. 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 1 or 2 did you sample with replacement and which did you sample without replacement?

This video provides a brief lesson on finding the probability of independent events.

Try It

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. Suppose that you sample four cards.

Which of the following outcomes are possible for sampling without replacement?

Which of the following outcomes are possible for sampling with replacement?


Probability of compound events

The probability of compound events combines at least two simple events, either the union of two simple events or the intersection of two simple events.

The probability that a coin will show head when you toss only one coin is a simple event. However, if you toss two coins, the probability of getting 2 heads is a compound event because once again it combines two simple events.

Suppose you say to a friend, " I will give you 10 dollars if both coins land on head."

Let's see what happens when your friend toss two coins:

If heads = H and tails = T, the different outcomes are HH, HT, TH, or TT.

As you can see, out of 4 possibilities, only 1 will give you HH.

Your friend has 25% chance of getting 10 dollars since one-fourth = 25%.

The example above is a good example of independent events. What are independent events?

When the outcome of one event does not affect the outcome of another event, the two events are said to be independent.

In our example above, when you toss two coins, neither coin has the power to influence the other coin.

This compound event is independent then. When two events are independent, you can use the following formula.

probability(A and B) = probability(A) × probability(B)

Let's use this formula to find the probability of getting 2 heads when two coins are tossed.


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Dependent Events

Experiment 1: A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack?

Analysis: The probability that the first card is a queen is 4 out of 52. However, if the first card is not replaced, then the second card is chosen from only 51 cards. Accordingly, the probability that the second card is a jack given that the first card is a queen is 4 out of 51.

Conclusion: The outcome of choosing the first card has affected the outcome of choosing the second card, making these events dependent.

Definition: Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.

Now that we have accounted for the fact that there is no replacement, we can find the probability of the dependent events in Experiment 1 by multiplying the probabilities of each event.

Experiment 1: A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack?

P(queen on first pick) = 4
52
P(jack on 2nd pick given queen on 1st pick) = 4
51
P(queen and jack) = 4 · 4 = 16 = 4
52 51 2652 663

Experiment 1 involved two compound, dependent events. The probability of choosing a jack on the second pick given that a queen was chosen on the first pick is called a conditional probability.

The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A) [pronounced as The probability of event B given A].

The notation used above does not mean that B is divided by A. It means the probability of event B given that event A has already occurred. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson.

Multiplication Rule 2: When two events, A and B, are dependent, the probability of both occurring is:

Let's look at some experiments in which we can apply this rule.

Experiment 2: Mr. Parietti needs two students to help him with a science demonstration for his class of 18 girls and 12 boys. He randomly chooses one student who comes to the front of the room. He then chooses a second student from those still seated. What is the probability that both students chosen are girls?

Probabilities P(Girl 1 and Girl 2) = P(Girl 1) and P(Girl 2|Girl 1)

= 18 · 17
30 29
= 306
870
= 51
145

Experiment 3: In a shipment of 20 computers, 3 are defective. Three computers are randomly selected and tested. What is the probability that all three are defective if the first and second ones are not replaced after being tested?

Probabilities: P(3 defectives) =

3 · 2 · 1 = 6 = 1
20 19 18 6840 1140

Experiment 4: Four cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing a ten, a nine, an eight and a seven in order?

Probabilities: P(10 and 9 and 8 and 7) =

4 · 4 · 4 · 4 = 256 = 32
52 51 50 49 6,497,400 812,175

Experiment 5: Three cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing 3 aces?

4 · 3 · 2 = 24 = 1
52 51 50 132,600 5,525

Summary: Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed. The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A). When two events, A and B, are dependent, the probability of both occurring is: P(A and B) = P(A) · P(B|A)

Exercises

Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.


Rule of Sum for Mutually Exclusive Events

Now think about whether or not it is possible to roll a number that is both odd and composite. As it turns out, it is not possible in the sample space of six-sided dice rolls to roll an odd composite number. The events O O O and C C C are called mutually exclusive, meaning they cannot both happen at the same time.

This distinction is very important for the rule of sum for mutually exclusive events:

" ∪ cup ∪ " is the symbol for a union. Because events are sets, unions of events can be understood in much the same way as unions of sets. P ( A ∪ B ) P(Acup B) P ( A ∪ B ) is the probability of either event A A A or event B B B happening.

This rule can be intuitively understood with a Venn diagram showing the sample space which includes events A A A and B B B :

Let S S S be a sample space which includes mutually exclusive events A A A and B B B . The Venn diagram of this sample space is pictured below:

Note that there is no overlap between events A A A and B B B . When events are mutually exclusive, it is not possible for both to happen at the same time.

The union of A A A and B B B is shown in blue. The probability of this union can be computed as follows:

P ( A ∪ B ) = ∣ A ∣ + ∣ B ∣ ∣ S ∣ = ∣ A ∣ ∣ S ∣ + ∣ B ∣ ∣ S ∣ = P ( A ) + P ( B ) . P(Acup B)=dfrac<|A|+|B|><|S|>=dfrac<|A|><|S|>+dfrac<|B|><|S|>=P(A)+P(B). P ( A ∪ B ) = ∣ S ∣ ∣ A ∣ + ∣ B ∣ ​ = ∣ S ∣ ∣ A ∣ ​ + ∣ S ∣ ∣ B ∣ ​ = P ( A ) + P ( B ) .

On a table, there are a total of 30 distinct books: 9 math books, 10 physics books, and 11 chemistry books.

What is the probability of getting a book that is not a math book?


Review for Midterm Two

Know how to compute the mean, the variance, and the standard deviation of a discrete random variable.

Chapter Five

Know the definitions of and.

Know when the binomial distribution applies and how to use it.

Know how to use the formulas

Know what the central limit theorem is.

Chapter Six

Know how to compute a confidence interval.

Know how to formulate a null hypothesis, an alternative hypothesis.

Know how to do a significance test.

Sample Problems

1. The original simple form of the Connecticut state lottery awarded the following prizes for each 100,000 tickets sold. The winners were chosen by drawing tickets at random.

Prize AmountNumber Awarded
1$5000
18$200
120$25
270$20

If you hold one ticket in this lottery, what is your probability of winning anything? What is the mean amount of your winnings?

Solution: The probability of winning anything is

The mean amount of winnings is

2. People with type O-negative blood are universal donors. That is, any patient can receive a transfusion of O-negative blood. Only 7% of the American population have O-negative blood. If 10 people appear at random to give blood, what is the probability that at least one of them is a universal donor?

Solution: The probability that at least one is a universal donor can be computed by using the complement rule.

Since the probability of being a universal donor is 0.07, the probability of being non-universal is 0.93. Thus

3. According to a market research firm, 52% of all residential telephone numbers in Los Angeles are unlisted. A telemarketing company uses random digit dialing equipment that dials residential numbers at random, regardless of whether they are listed in the telephone directory. The firm calls 500 numbers in Los Angeles. What is the exact distribution of the number X of unlisted numbers that are called? Use a suitable approximation to calculate the probability that at least half the numbers called are unlisted.

Solution: Since there are only two outcomes and the 500 numbers are selected independently of each other, the binomial distribution applies in this situation. The exact distribution for the number of unlisted numbers called is B(500,0.52)(k). Since n is fairly large, we can approximate the binomial distribution with a normal distribution where

The probability that at least half of the numbers called are unlisted is

To compute this probability from the normal distribution, we have to convert X = 250 to a z-score.

According to the normal curve table, approximately 0.185 of the curve is below this z. Thus about 81.5% of the curve is above that z, and we have an 81.5% chance of at least half the calls being to unlisted numbers.

4. Patients with chronic kidney failure may be treated by dialysis, using a machine that removes toxic wastes from the blood, a function normally performed by the kidneys. Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are the data for one patient (in milligrams of phosphorus per deciliter of blood):

The measurements are separated in time and can be considered to be an SRS of the patient's blood phosphorus level. Assuming that this level varies normally with mg/dl, give a 90% confidence interval for the mean blood phosphorus level. The normal range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that the patient has a mean phosphorus level that exceeds 4.8?

Solution: The six values reported here have a mean value of

The confidence interval is

where z* is chosen so that 90% of the normal curve lies between -z* and z*. Table D on page T-11 lists values of z* for various confidence levels C, and says that for C = 90%, z* = 1.645. Thus the 90% confidence interval is

If we formulate an alternative hypothesis that the patient's is greater than 4.8, we can compute a significance level by asking "if , what is the probability that we would measure by pure chance?" To compute this probability, we compute a z-score for the observed mean.

According to the normal probability table, 1 - 0.9382 = 0.0618 of the curve lies above this z. This is not significant at the usual 5% level.


Watch the video: Ανεξάρτητα ενδεχόμενα. Τυχαία Μεταβλητή ΤΜ. Κατανομή Πιθανότητας ΚΠ (September 2021).