Question

A box has eight balls. There are five green balls and three yellow. The five green balls are numbered 1 ,2,3,4 ,and 5. The three yellow balls are numbered 1,2 ,and 3. The box is shaken.

Answer the following questions, using fractions in p/q format or decimal values accurate to the nearest 0.01 . Here G will mean that a green ball is selected, Y that a yellow ball is selected, and E that an even- numbered ball is selected.

a)a (25%) Select the color/number pairs which comprise the sample

Gl G2 G3 G4 G5 G6 G7 G8 Yl Y2 Y3 Y4 Y5 Y6 (15%)

b)P(G) = C. (15%)

c)P (GE)= d (1596)

d)P(Gr E)= e (1596)

e)P(G U E) = f. (1596) Are G and E mutually exclusive? (pick one)

Answer 1

Answer:

a) G1, G2, G3, G4, G5, Y1, Y2, Y3

b) P(G) = 5/8

c) P(G|E) = 2/3

d) P(G∩E) = 1/4

e) P(G∪E) = 3/4

G and E are not mutually exclusive

Step-by-step explanation:

Data:

Green Balls = 5 (Numbered 1,2,3,4,5)

Yellow Balls= 3 (Numbered 1,2,3,)

Total Number of Balls = 8

a) The sample space comprises of 5 Green (G) balls numbered 1,2,3,4,5 and 3 Yellow (Y) balls numbered 1,2,3.

So the sample space comprises of:

G1, G2, G3, G4, G5, Y1, Y2, Y3

b) The probability that a green ball is selected can be computed by:

P(G) = Number of Green Balls / Total Number of Balls

P(G) = 5/8

c) P(G|E) = P(G∩E)/P(E)

To calculate P(G∩E), we need to count the number of balls which are green AND even numbered. These are: G2 and G4. The total number of green balls are 8.

So, P(G∩E) = Number of  green and even balls / Total Number of Balls

      P(G∩E) = 2/8

To calculate P(E), we know that the number of even balls are 3.

P(E) = Number of even balls/Total number of balls

P(E) = 3/8

Therefore,

P(G|E) = (2/8) / (3/8)

P(G|E) = 2/3

d) From the previous part, P(G∩E) = 2/8

It can also be calculated using the conditional probability formula:

P(G|E) = P(G∩E) / P(E)

P(G∩E) = P(E) x P(G|E)

            = (3/8) x (2/3)

P(G∩E) = 2/8 = 1/4

e)P(G∪E) = P(G) + P(E) - P(G∩E)

               = (5/8) + (3/8) - (2/8)

  P(G∪E) = 6/8 = 3/4

For mutually exclusive events, P(AUB) = P(A) + P(B)

Here, P(GUE) = 6/8 , P(G) = 5/8 and P(E) = 3/8

P(GUE) = P(G) + P(E)

6/8 = 5/8 + 3/8

6/8 ≠ 8/8

So, events G and E are NOT Mutually Exclusive