Question

In: Statistics and Probability

suppose he roll two dice one black and one red you record the outcomes in order...

suppose he roll two dice one black and one red you record the outcomes in order for example the outcome (35) notes at three on the black die And a five on the red dye. What a beauty event that the black guy is even, B be the event at the red dye is odd, and C be the event that the dice sum to 10

(a) List and describe the outcomes in the sample space S and the events A, B and C. report the number of outcomes in each set

(b) find P(A), P(B), P(C)

(d) fine P(A and C). are A and C independent

Solutions

Expert Solution

Two dies are rolled simultaneously, thus, the sample space S is given by

S = {( 11 ),( 12 ),( 13 ),( 14 ),( 15 ),( 16 ),( 21 ),( 22 ),( 23 ),( 24 ),( 25 ),( 26 ),( 31 ),( 32 ),( 33 ),( 34 ),( 35 ),( 36 ),( 41 ),( 42 ),

( 43 ),( 44 ),( 45 ),( 46 ),( 51 ),( 52 ),( 53 ),( 54 ),( 55 ),( 56 ),( 61 ),( 62 ),( 63 ),( 64 ),( 65 ),( 66 )}

#S = number of observations in sample space = 36

where, ( i j ) : i on black die and j on red die

The events are

A : Even number occurred on black die ; in ( i j ) i is even

A = {( 21 ),( 22 ),( 23 ),( 24 ),( 25 ),( 26 ),( 41 ),( 42 ),( 43 ),( 44 ),( 45 ),( 46 ),( 61 ),( 62 ),( 63 ),( 64 ),( 65 ),( 66 )}

#A = 18

B : Odd number occurred on red die ; in ( i j ) j is odd

B = {( 11 ),( 13 ),( 15 ),( 21 ),( 23 ),( 25 ),( 31 ),( 33 ),( 35 ),( 41 ),( 43 ),( 45 )( 51 ),( 53 ),( 55 ),( 61 ),( 63 ),( 65 )}

#B = 18

C : Sum of numbers on both dies is 10 ; ( i j ) is such that i + j = 10

C = {( 46 ),( 55 ),( 64 )}

#C = 3

P(A) = #A / #S = 18/36 = 1/2 = 0.5

P(B) = #B / #S = 18/36 = 1/2 = 0.5

P(C) = #C / #S = 3/36 = 1/12

(A and B) = {( 21 ),( 23 ),( 25 ),( 41 ),( 43 ),( 45 ),( 61 ),( 63 ),( 65 )} ; #(A and B) = 9

P(A and B) = #(A and B) / #S = 9/36 = 1/4 = 0.25

P(A)*P(B) = 0.5 * 0.5 = 0.25

Since, P(A and B) = P(A)*P(B) , Events A and B are INDEPENDENT

(A and C) = {( 46 ),( 64 )} ; #(A and C) = 2

P(A and C) = #(A and C) / #S = 2/36 = 1/18

P(A)*P(C) = (1/2) * (1/12) = 1/24

Since, P(A and C) is not equal to P(A)*P(C) , Events A and C are NOT INDEPENDENT

orchestra answered 1 year ago

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