Statistics with Probability

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  • #12969

    Ezra Halleck
    Participant

    1. The probability that an unfair coin lands on heads is 0.6. Coin is tossed 3 times.
    a. List the sample space.
    b. Construct a random variable X which counts number of heads for the three tosses.
    c. Find P(X<1), E(X) and E(1/(X+1))

    #16067

    Anonymous
    Inactive

    Is this the sample space? {HHH, THH, HTH, HHT, THT, TTH, HTT, TTT}.

    #16203

    mendozak
    Member

    a.

    S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

    P (HHH) = .216
    P (HHT) = P (HTH) = P (THH) = .144
    P (HTT) = P (THT) = P (TTH) = .096
    P (TTT) = .064

    .216 + (3).144 + (3).096 + .064 = 1

    b.

    P (X = 3) = .216
    P (X = 2) = .432
    P (X = 1) = .288
    P (X = 0) = .064

    .216 + .432 + .288 + .064 = 1

    c.

    P (X<1) = P (X=0) = .064
    E (X) = (3*(.216)) + (2*(.432)) + (1*(.288)) + (0*(.064)) = 1.8

    #16211

    Ezra Halleck
    Participant

    Good work! From the distribution, we can see a shifting of the middle towards the right (and skewing to the left). With a fair coin, the mean would be 1.5 rather than 1.8.

    #16214

    Rob
    Member

    Visual representation of the problem as a tree

    #16215

    Ziyuan Wu
    Member

    For the second part of (c), E(1/(X+1))

    E(1/(X+1))=0.064+(0.288/2)+(0.432/3)+(0.216/4)=0.406

    #16220

    Rob
    Member

    sorry for the rough sketch quality

    #16237

    well looking at these solution from other persons are really helping to figure out where i went wrong. NICE WORK mendozak

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