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) <u>P(G|E) = P(G∩E)/P(E)</u>
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
<u>P(G∩E) = 2/8 </u>
To calculate P(E), we know that the number of even balls are 3.
P(E) = Number of even balls/Total number of balls
<u>P(E) = 3/8</u>
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