Question

In: Statistics and Probability

Section I You toss a coin and roll a die simultaneously. If the coin shows heads,...

Section I
You toss a coin and roll a die simultaneously. If the coin shows heads, the experiment outcome is equal to the value shown on the die. If the coin shows tails, the experiment outcome is equal to twice the value shown on the die. Assume that the coin and the die are fair. Let ? be 1 if the coin shows heads and 2 if the coin shows tails, ?be the outcome of rolling the die, and ? the outcome of the experiment. Notice that ?, ?, and ? are random variables.

  1. What is the minimal sample space in this experiment? (4 points)

  2. Identify the event that the outcome of the experiment is, at least, 10. (4 points)

  3. What is the value of ??(? ≤ 9)? (4 points)

  4. Whatisthevalueof??(4≤?≤10|?=2)?(4points)

  5. Whatisthevalueof??(?=2|4≤?≤10)?(4points)

Solutions

Expert Solution

X , Y and Z can take following values

X Y Z
1 1 1
2 2
3 3
4 4
5 5
6 6
2 1 2
2 4
3 6
4 8
5 10
6 12

Thus minimal sample space consists of 12 points given by

S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),(2,8),(2,10),(2,12)}

The sample space consists of points of (X,Z)

The event for the outcome 10 is X=2, Y=5

The event for the outcome 12 is X=2, Y=6  

Thus two events correspond to outcome of at least 10

To find P(Z 9)

From sample space , there are 10 events correspond to outcome Z 9

total number of events in the sample space = 12

P(Z 9) = 10/12 = 0.8333

P(Z 9) =0.8333

To find P(4Z 10 I X= 2)

P(4Z 10 I X= 2) = P( 4Z 10 , X= 2) / P( X=2)

From sample space , there are 4 events correspond to 4Z 10 , X= 2

total number of events in the sample space = 12

P( 4Z 10 , X= 2) = 4/12

P(X=2) = 6/12

P(4Z 10 I X= 2) = 4/6 = 0.6667

P(4Z 10 I X= 2) =0.6667

To find P( X=2 I 4Z 10) = ?

Using Bayes' theorem

P( X=2 I 4Z 10) = P(4Z 10 I X= 2) *P(X=2) / P(4Z 10)

P(4Z 10 I X= 2) =0.6667

P(X=2) = 0.5

There are 7 events that correspond to the outcome 4Z 10 ( highlighted below)

P(4Z 10) = 7/12 =0.5833

Therefore , P( X=2 I 4Z 10) = 0.6667*0.5 / 0.5833= 0.5715

P( X=2 I 4Z 10) =0.5715

X Y Z
1 1 1
2 2
3 3
4 4
5 5
6 6
2 1 2
2 4
3 6
4 8
5 10
6 12

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