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Sedbober [7]
3 years ago
7

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The

three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
List:
1) simple space,
2) Enter probability P(G) as a fraction
3) P(G/E)
4) P(G AND E)
5) P(G or E)
6) Are G and E are mutually exclusive
Mathematics
1 answer:
Marina86 [1]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

G = card drawn is green

Y = card drawn is yellow

E = card drawn is even-numbered

List:

Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}

2) P(G) = 5/8

3) P(G/E) = P(GE)/P(E)

GE = {G2, G4}

Hence P(G/E) = 2/5

4) GE = {G2, G4}

P(GE) = 2/8 = 1/4

5) P(G or E) = P(G)+P(E)-P(GE)

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

6) No there is common element as G2 and G4

Cannot be mutually exclusive

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