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

in progress 0
Delilah 3 months 2022-02-10T06:40:32+00:00 1 Answer 0 views 0

Answers ( )

    0
    2022-02-10T06:41:51+00:00

    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

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )