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    I have a test coming up for my Statistics class, and my professor put up last year's test on the course website so we could see what kind of questioins are coming up. I've tried solving them, and I want to see if I got them right. Can anyone check my solutions and answer and see they're correct?

    I. Poker
    A player draws five cards randomly from an ordinary deck of 52 cards.

    1. What is the probability that he gets full house, that is, three cards of one denomination and two cards of a second denomination? For example: Three kings and two 8's.

    I did 13C1*4C3*12C1*4C2/52C5=0.00144...

    2. What is the probability that he gets three of a kind, that is, three cards of one denomination, a fourth card of a second denomination, and a fifth card of a third denomination. For example: three 7's, an 8 and a King.

    I did 13C1*4C3*12C2*4C1*4C1/52C5=0.0211...

    3. What is the probability that he gets at least one card from each suit? For example: a 3 of heart, a 4 of spade, an 8 of club, a king of club and a 10 of diamond.

    I did 13C2*13C1*13C1*13C1*4C1/52C3=0.26374

    II. Chess Tournament
    Bill and John play a chess tournament. John is more skilled than Bill and the probability of John winning a game is twice as that of Bill. A player is a winner until he wins 2 games in a row or 3 games altogether.

    1. In what percentage of all possible cases does the tournament end because John wins 3 games with winning 2 in a row?

    John has a 2/3 chance of winning and Bill has a 1/3 chance.
    The instances in which John wins 3 games with 2 in a row is...
    JBJJ and BJBJJ
    JBJJ becomes 2/3*1/3*2/3*2/3=8/81
    BJBJJ becomes 1/3*2/3*1/3*2/3*2/3=8/243
    Add them together to get 32/243

    2. In what probability does the tournament end because Bill wins 3 games without winning 2 in a row?

    The instance in which Bill wins 3 games without winning 2 in a row is...
    BJBJB
    BJBJB becomes 1/3*2/3*1/3*2/3*1/3=4/243

    3. Knowing that the tournament ends because Bill wins 3 games altogether, in what is the probability that Bill wins 2 games in a row?

    The instances in which the tournament ends becaues Bill wins 3 games altogether are
    BJBJB and BJBB and JBJBB
    BJBJB is 4/243
    BJBB is 1/3*2/3*1/3*1/3=2/81
    JBJBB is 2/3*1/3*2/3*1/3*1/3=4/243
    The probability of the tournament ending because Bill wins 3 games altogether are 4/243+6/243+4/243=14/243
    From that, the probability of tournament ending because Bill wins 3 games altogether with 2 in a row is BJBB and JBJBB, which becomes 10/243
    (10/243)/(14/243)=10/14=0.71428

    III. Urns and Balls
    Urn I contains 5 white balls and 7 black balls. Urn II contains 4 white balls and 4 black balls. Urn III contains 10 white balls and no black balls. An urn is selected randomly and a ball drawn randomly from it is observed to be black and then returned to the same urn.

    1. What is the probability that the selected urn is urn I?

    Considering that the urn selected has a black ball in it, I'd say the chance is 1/2, or 0.5, since only two urns (Urn I and Urn II) have black balls in them.

    2. If a second ball is drawn from the same urn, what is the probability that it is a white ball?

    If Urn I was chosen, then 1/2*7/12*5/12=35/288
    If Urn II was chosen, then 1/2*4/8*4/8=1/8
    Add them together to get 35/288+18/288=53/288=0.184

    If you see anything wrong with my calculations, please tell me what it was that I did.
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    I think you did something wrong with Part III (haven't checked the other two yet):

    1. P(Urn One| Black) = P(Urn One AND Black)/P(Black) = (1/3*7/12)/{(1/3*7/12) + (1/3 + 1/2)} = 7/13 (a bit more than 0.5)

    2. Urn I = 7/13 * 5/12 = 35/158

    Urn II = 6/13 * 4/8 = 3/13

    Total probability of getting white ball = 0.452

    Which makes sense if you think about it. You get a black ball. So it's either Urn I and II. The probability of getting a white ball after that is just less than half. 0.184 is obviously too small.
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    Ooh, thanks for that. I forgot all about the P(A l B) stuff. I'll have to reread that segment.
 
 
 
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