Suppose that you have 11 cards. 5 are green and 6 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 6 yellow cards are numbe
Question
Suppose that you have 11 cards. 5 are green and 6 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 6 yellow cards are numbered 1, 2, 3, 4, 5, and 6. 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
a. List the sample space.
b. Enter probability P(G) as a fraction
c. P(G/E)
d. P(G AND E)
e. P(G or E)
f. Are G and E are mutually exclusive
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2022-02-10T06:40:32+00:00
2022-02-10T06:40:32+00:00 1 Answer
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Answer:
a. S={G1,G2,G3,G4,G5,Y1,Y2,Y3,Y4,Y5,Y6}
b. P(G)=5/11
c. P(G/E)=2/5 or 0.4
d. P(G and E)=2/11 or 0.1818
e. P(G or E)=8/11 or 0.7273
f. G and E not mutually exclusive
Step-by-step explanation:
G={G1,G2,G3,G4,G5}
n(G)=5
Y={Y1,Y2,Y3,Y4,Y5,Y6}
n(Y)=6
E={G2,G4,Y2,Y4,Y6}
n(E)=5
a)
The possible outcomes in sample space are
S={G1,G2,G3,G4,G5,Y1,Y2,Y3,Y4,Y5,Y6}
b)
P(G)=n(G)/n(S)=5/11
c)
P(G/E)=P(G and E)/P(E)
G and E={G2,G4}
n(G and E)=2
P(G and E)=n(G and E)/n(S)=2/11
P(E)=n(E)/n(S)=5/11
P(G/E)=2/5
d)
G and E={G2,G4}
n(G and E)=2
P(G and E)=n(G and E)/n(S)=2/11
e)
P(G or E)= P(G)+P(E)-P(G and E)
P(G or E)=5/11+5/11-2/11
P(G or E)=8/11
f)
G and E are not mutually exclusive because G and E have 2 outcomes in common.
G and E={G2,G4}
Thus, P(G and E)≠0 and G and E are not mutually exclusive events