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31 December, 08:24

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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Answers (1)
  1. 31 December, 11:39
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    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
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