Respuesta :
Answer:
i) X 2 3 4 5 6 7 8 9 10 11
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/26 2/36
X 12
P(X) 1/36
And if we check for this case when we add the probabilities we got:
[tex] \frac{1}{36} +\frac{2}{36}+\frac{3}{36} +\frac{4}{36} +\frac{5}{36} +\frac{6}{36} +\frac{5}{36} +\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36} =1[/tex]
ii) [tex] P(X \geq 8) =\frac{5}{36} +\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{15}{36} [/tex]
iii)[tex] E(X) =2*\frac{1}{36} +3*\frac{2}{36}+4*\frac{3}{36} +5*\frac{4}{36} +6*\frac{5}{36} +7*\frac{6}{36} +8*\frac{5}{36} +9*\frac{4}{36}+10*\frac{3}{36}+11*\frac{2}{36}+12*\frac{1}{36} =\frac{252}{36}=7[/tex]Explanation:
For this case we can define the following sample space for a two balanced dice:
(1,1) (1,2) (1,3) (1.4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
We have a total of 36 possible pairs
Part a
(i) Obtain the probability distribution of X (i.e. what are the possible values for X and the probability for obtaining each value?).
And the possible values ofr X on this case are : X = 2,3,4,5,6,7,8,9,10,11,12
Now we can obtain the porbability distribution for X like this:
X 2 3 4 5 6 7 8 9 10 11
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/26 2/36
X 12
P(X) 1/36
And if we check for this case when we add the probabilities we got:
[tex] \frac{1}{36} +\frac{2}{36}+\frac{3}{36} +\frac{4}{36} +\frac{5}{36} +\frac{6}{36} +\frac{5}{36} +\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36} =1[/tex]
(ii) What is the probability for obtaining X >= 8?
For this case we want this probability:[tex] P(X \geq 8) = P(X=8)+P(X=9)+P(X=10)+P(X=11)+P(X=12)[/tex]And if we replace we got:
[tex] P(X \geq 8) =\frac{5}{36} +\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{15}{36} [/tex]
(iii) What is the average value of X?
For this case we can use the definition of expected value like this and we got:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) [/tex]
And if we replace we have:
[tex] E(X) =2*\frac{1}{36} +3*\frac{2}{36}+4*\frac{3}{36} +5*\frac{4}{36} +6*\frac{5}{36} +7*\frac{6}{36} +8*\frac{5}{36} +9*\frac{4}{36}+10*\frac{3}{36}+11*\frac{2}{36}+12*\frac{1}{36} =\frac{252}{36}=7[/tex]
