Question

For each part of this question, complete the statement or statements by typing in your answers in the provided spaces:

1. If the probabilities: that event A occurs is .55; that event B occurs is .60; and, that both A and B occur is .35,

The probability (up to 4 decimal places if needed) that events A or B (or both) occur is ______

The probability (up to 4 decimal places if needed) that event B occurs given that A occurs is ______          

2. If the probabilities: that event A occurs is .40; that event B occurs is .7; and, that event B occurs given that event A occurred is .4,

Complete the following contingency table by typing the appropriate probabilities (to 2-decimal places in the spaces provided

                                                      A                   A^C              total

                                    B           ____           ____           ____                     

                                    B^C         ____            ____            ____

                              Total        ____           ____            ____

c. Given the following probability distribution of X:

x                P(x)

0                .15

1                .20

2                .40

3                .25

                  The expected value (to 2-decimal places) of X is _____

                  If Y = 3 + 4X, the expected value (to 2 decimal places) of Y is _____

Answer 1

(a)

(i)

P(A) = 0.55

P(B) = 0.60

P(AB) = 0.35

So,

The probability (up to 4 decimal places if needed) that events A or B (or both) occur is = P(A+B) = P(A) + P(B) - P(AB)

= 0.55 +0.60 - 0.35

= 0.8000

So,

Answer is:

0.8000

(ii)

The probability (up to 4 decimal places if needed) that event B occurs given that A occurs is = P(B/A) = P(AB)/ P(A)

= 0.35/0.55

= 0.6364

So,

Answer is:

0.6364

(b)

(i)

Given:
P(A) = 0.40

P(B) = 0.70

P(B/A) = 0.4

So,

P(AB) = P(B/A) X P(A)

        = 0.4 X 0.4

        = 0.16

(ii)

The contingency table is filled up as follows:

	A	AC	Total
B	0.16	0.54	0.70
BC	0.40 - 0.16 = 0.24	0.06	0.30
Total	0.40	0.60	1.00

(c)

(i)

From the given data, the following Table is calculated:

x	p	x p
0	0.15	0
1	0.20	0.20
2	0.40	0.80
3	0.25	0.75
Total

So,

  The expected value (to 2-decimal places) of X is 1.75

(ii)
E(Y) = 3 + 4 E(X) = 3 + (4 X 1.75) = 10

So,

Answer is:

10