Intro a probability (4/6)

7/30/2019 Intro a probability (4/6)

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How to get started on a solution?

Ani Adhikari and Philip Stark Statistics 2.2X Lecture 1.4 1 / 7
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There is no Magic Method, and very little plug and chug.

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Each problem requires its own combination of the two main rules.



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No matter what method you use, its going to be important to



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No matter what method you use, its going to be important to be precise in your use of terminology and notation



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pay attention to detail

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avoid rushing to conclusions

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avoid rushing to conclusions

Here are a few efficient ways to approach problems.

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Count the number of trials

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trial: one stage of a process that involves randomness

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One trial



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One trial

If you have the list of all possible outcomes, try just counting how manyof those outcomes satisfy the conditions of the event.

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One trial


Example 1. A random number generator picks at random from the tendigits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Find the chance that the digit it picksis prime and greater than 5.

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One trial


Example 1. A random number generator picks at random from the tendigits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Find the chance that the digit it picksis prime and greater than 5.

Solution. Only one outcome works: the digit 7. Answer = 1/10

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One trial: summarized data

One trial (continued) If you dont have the whole list but just summaries of it, the rules willhelp. Drawing Venn diagrams helps too.



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Example 2. In a class, 15% of the students have read Hamlet and 40%

have read The Merchant of Venice; 5% of the students have read both.What is the chance that a student picked at random has read one of theplays but not the other?



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Solution. List the ways. That is, partition the event into its componentpieces.



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A: Hamlet but not Merchant of Venice: 10%B: Merchant of Venice but not Hamlet: 35%



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A: Hamlet but not Merchant of Venice: 10%B: Merchant of Venice but not Hamlet: 35%

Add these up. Answer: 45%.

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Two trials: table of outcomes

Two trials, modest number of outcomes

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Two trials, modest number of outcomes Draw a table of all possible outcomes.

Example 3. In my pocket I have a quarter, two dimes, and a nickel. If Ipull out two coins at random (without replacement), what is the chancethat I pull out more than 15 cents?

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Solution.

Coin ICoin II Q D1 D2 N

Q

D1 D2

N

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Solution.

Coin ICoin II Q D1 D2 N

Q

D1 D2

N

8 out of 12 equally likely cells. Answer 8/12 = 2/3

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The crucial first trial

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More than two trials,

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More than two trials, or just two trials but each with many outcomes

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Ask, What does the first trial have to be?



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Ask, What does the first trial have to be?If you have a clear answer, youre on your way to a solution.


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Example 4. A die has six equally likely faces. The faces show 1, 2, 3, 4,5, and 6 spots respectively. The die is rolled 5 times. Find the chance thatnone of the rolls shows a multiple of 3 spots.


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Solution. The first roll has to show 1, 2, 4, or 5 spots.


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Solution. The first roll has to show 1, 2, 4, or 5 spots. So does the

second; so does the third; and so on.


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second; so does the third; and so on.(4/6)


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second; so does the third; and so on.(4/6)


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second; so does the third; and so on.(4/6) (4/6) (4/6) (4/6) (4/6) = (4/6)5 = 13.17%


When its not clear what the first trial must be
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When it s not clear what the first trial must be

Example 5. A drawer contains 10 red socks and 10 blue socks. I pull outtwo socks at random without replacement. What is the chance they are ofthe same color?


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Solution. Theres no restriction on the first sock.


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Solution. Theres no restriction on the first sock. No matter what thecolor of the first sock, the chance that the second one has the same color

is 9/19.


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is 9/19. Answer: (20/20) (9/19) = 9/19 = 47.37%


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is 9/19. Answer: (20/20) (9/19) = 9/19 = 47.37%

Example 6. Repeat Example 5, if the drawer has 4 red socks and 16 bluesocks.


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is 9/19. Answer: (20/20) (9/19) = 9/19 = 47.37%


Solution. List the ways.


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is 9/19. Answer: (20/20) (9/19) = 9/19 = 47.37%


Solution. List the ways.RR: (4/20) (3/19) = 0.0316BB: (16/20) (15/19) = 0.6316


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is 9/19. Answer: (20/20) (9/19) = 9/19 = 47.37%


Solution. List the ways.RR: (4/20) (3/19) = 0.0316BB: (16/20) (15/19) = 0.6316

Add these up. Answer: 66.32%


at least one
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at least one

When its not clear what the first trial must be (continued)


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t t


Example 7. A poker hand (5 cards) is dealt from a well shuffled deck.What is the chance that there is at least one ace in the hand?


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Solution. complement of at least one:


at least one


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Solution. complement of at least one: none


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P(at least one ace)

= 1 P(no aces)


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P(at least one ace)

= 1 P(no aces)

= 1 (48/52) (47/51) (46/50) (45/49) (44/48)

= 34.11%

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Intro a probability (4/6)

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