the probability the probability that the knoit modest prizes and would the if yo

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the probability the probability that the knoit modest prizes and would the if you quit now or if curtain O. He then behind curtain B? Car CASE 6 then be onered the opportunity will choice 2. After Monty shows risk those prizes you like he your you suppose that you contestant ie, you can keep behind curtain C, what on this show area curtain A or you do? probability that the the Monty has just given of What do A What is you a free trip touring ravic waste in Ba that the car is behind sites around the country He now answer that question, To help you questions. offers you a trade: Give up the trip answering these first in exchange for a gamble. On the try approxim ity of be To Bunt or Not to Bunt, That CASE 6.2 CA. Is the Question o Sport generates as many Winter 1997) offers baseball lovers statistics as baseball. another opportunity to analyze Reporters managers, and numbers associated with the game. example, the probability of scoring fans argue and discuss strategies on Table 1 lists the probabilities of the basis of these statistics An article scoring at least one run in situations at least one run when there are no in Chance A Statistician Reads the that are defined by the number of outs and a man is on first base is 39 Sports Page," Hal S. Stern, Vol. 1, outs and the bases occupied. For If the bases are loaded with one out,

Explanation / Answer

6.2 To Bunt or Not to Bunt :

Table 1: Probability of scoring answers

Outcome

Probability

Bases Occupied

outs

Probability of scoring

Joint Probability

P(Outcome1)

.75

2nd

.42

.3150

P(Outcome2)

.10

1st

.26

.0260

P(Outcome3)

.10

none

.07

.0070

P(Outcome4)

.05

1st and 2nd

.59

.0295

Total probability of scoring runs = .3775

We wil consider for outcome 1

So here given probability =0.75

Given in question as the bunt is successful .the advances one base.and the batter is out .so probability may occupied as 2nd place

So the probability for p(1) = given probability probability of scoring in base

= 0.750.42

= 0.3150

The probability for p(2) = 0.100.26

= 0.260

The probability for p(3) = 0.10 X 0.07

= 0.070

The probability for p(4) = 0.5 0.59

= 0.295

Total scoring runs p = 0.3775

Now to Bunt or Not :

The probability of scoring with a runner on first base with no outs = (.39)

The probability after Bunting is =0.3775

So in case above runner on first base is greater than bunting .so now “NOT TO BUNT”

Case 6.3:

Should He Attempt To steal a Base

For each of the possible number of outs (0,1,2). Deteremine for second Base

0 outs:

Probability of scoring any runs from first base = 0.39

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) X (0.57) = 0.3876

Decision: “Do not attempt to steal”. As it was less than probability of scoring runs

1 out:

Probability of scoring any runs from first base = 0.26

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) (0.42) = 0.2856

Decision: “Attempt to steal” As it was greater .

2 outs:

Probability of scoring any runs from first base = 0.13

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) (0.24) = 0.1632

Decision: ” Attempt to steal”.

6.2 To Bunt or Not to Bunt :

Table 1: Probability of scoring answers

Outcome

Probability

Bases Occupied

outs

Probability of scoring

Joint Probability

P(Outcome1)

.75

2nd

.42

.3150

P(Outcome2)

.10

1st

.26

.0260

P(Outcome3)

.10

none

.07

.0070

P(Outcome4)

.05

1st and 2nd

.59

.0295

Total probability of scoring runs = .3775

We wil consider for outcome 1

So here given probability =0.75

Given in question as the bunt is successful .the advances one base.and the batter is out .so probability may occupied as 2nd place

So the probability for p(1) = given probability probability of scoring in base

= 0.750.42

= 0.3150

The probability for p(2) = 0.100.26

= 0.260

The probability for p(3) = 0.10 X 0.07

= 0.070

The probability for p(4) = 0.5 0.59

= 0.295

Total scoring runs p = 0.3775

Now to Bunt or Not :

The probability of scoring with a runner on first base with no outs = (.39)

The probability after Bunting is =0.3775

So in case above runner on first base is greater than bunting .so now “NOT TO BUNT”

Case 6.3:

Should He Attempt To steal a Base

For each of the possible number of outs (0,1,2). Deteremine for second Base

0 outs:

Probability of scoring any runs from first base = 0.39

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) X (0.57) = 0.3876

Decision: “Do not attempt to steal”. As it was less than probability of scoring runs

1 out:

Probability of scoring any runs from first base = 0.26

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) (0.42) = 0.2856

Decision: “Attempt to steal” As it was greater .

2 outs:

Probability of scoring any runs from first base = 0.13

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) (0.24) = 0.1632

Decision: ” Attempt to steal”.

6.2 To Bunt or Not to Bunt :

Table 1: Probability of scoring answers

Outcome

Probability

Bases Occupied

outs

Probability of scoring

Joint Probability

P(Outcome1)

.75

2nd

.42

.3150

P(Outcome2)

.10

1st

.26

.0260

P(Outcome3)

.10

none

.07

.0070

P(Outcome4)

.05

1st and 2nd

.59

.0295

Total probability of scoring runs = .3775

We wil consider for outcome 1

So here given probability =0.75

Given in question as the bunt is successful .the advances one base.and the batter is out .so probability may occupied as 2nd place

So the probability for p(1) = given probability probability of scoring in base

= 0.750.42

= 0.3150

The probability for p(2) = 0.100.26

= 0.260

The probability for p(3) = 0.10 X 0.07

= 0.070

The probability for p(4) = 0.5 0.59

= 0.295

Total scoring runs p = 0.3775

Now to Bunt or Not :

The probability of scoring with a runner on first base with no outs = (.39)

The probability after Bunting is =0.3775

So in case above runner on first base is greater than bunting .so now “NOT TO BUNT”

Case 6.3:

Should He Attempt To steal a Base

For each of the possible number of outs (0,1,2). Deteremine for second Base

0 outs:

Probability of scoring any runs from first base = 0.39

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) X (0.57) = 0.3876

Decision: “Do not attempt to steal”. As it was less than probability of scoring runs

1 out:

Probability of scoring any runs from first base = 0.26

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) (0.42) = 0.2856

Decision: “Attempt to steal” As it was greater .

2 outs:

Probability of scoring any runs from first base = 0.13

Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (0.68) (0.24) = 0.1632

Decision: ” Attempt to steal”.

Outcome

Probability

Bases Occupied

outs

Probability of scoring

Joint Probability

P(Outcome1)

.75

2nd

.42

.3150

P(Outcome2)

.10

1st

.26

.0260

P(Outcome3)

.10

none

.07

.0070

P(Outcome4)

.05

1st and 2nd

.59

.0295

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