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Another roll of the…

Posted by DP on March 18, 2011 – 11:01 am

This time we have only a single {fair} die and roll it exactly 5 times. What is the probability that you will roll exactly:

a) [2] 3’s?

b) [4] 6’s?

c) [0] 1’s?


This post is under “Tom” and has 9 respond so far.
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9 Responds so far- Add one»

  1. 1. cazayoux Said:

    First stab at it.

    a) 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6
    b) 1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6
    c) 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6

  2. 2. Chris Said:

    a) (1/6)^2 * (5/6)^4 * C(6,2) = 15*625/46656 ≈ 0.2009
    b) (1/6)^4 * (5/6)^2 * C(6,4) = 15*25/46656 ≈ 0.008038
    c) (5/6))^6 = 15625/46656 ≈ 0.3349

    where C(6,2) = C(6,4) = 6!/(4! 2!) = 15

  3. 3. DP Said:

    You may both consider looking back over the problem…

  4. 4. Chris Said:

    Ooops. 5 rolls, we did 6. (Thanks for fixing my typos, DP).

    a) (1/6)^2 * (5/6)^3 * C(5,2) = 10*125/7776 ≈ 0.1608
    b) (1/6)^4 * (5/6) * C(5,4) = 5*5/7776 ≈ 0.0032
    c) (5/6)^5 = 3125/7776 ≈ 0.4019

  5. 5. DP Said:

    Yep. Chris has it.
    This is an example of Binomial Probability.

    x: # of successes
    n: # of trials
    p: probability of success
    q: probability of failure ( = 1 – p )

    b(x;n,p) = C(n,x) * p^x * q^(n-x)
    where C(n,x) = P(n,x)/x!
    where P(n,x) = n!/(n-x)!

    so b(x;n,p) = n!/[(n-x)! x!] * p^x * q^(n-x)

    For all three cases we roll the die 5 times, so n=5.
    And because we are only looking for 1 specific numbre out of the 6, p=1/6 and q=5/6.

    for our problems, b(x;5,1/6) = 5!/[(5-x)! x!] * (1/6)^x * (5/6)^(5-x)

    a) when x=2; b(2;5,1/6) = 5!/(3! 2!) * (1/6)^2 * (5/6)^3 = 1250/7776 or 0.1607510…

    b) when x=4; b(4;5,1/6) = 5!/(1! 4!) * (1/6)^4 * (5/6)^1 = 25/7776 or 0.003215…

    c) when x=0; b(0;5,1/6) = 5!/(0! 5!) * (1/6)^0 * (5/6)^5
    since 0!=1 and p^0=1, we get 5!/5! * 1 * (5/6)^5 = 25/7776 or 0.40187757…

    I can’t figure out how to make the “roughly equal to” symbol like Chris has. Oh well, you get the point.

  6. 6. DP Said:

    cazayoux,
    I had to do some brushing up on my probability & statistics on this one to figure out why to solve it this way. I would have probably tried it exactly how you did had someone else posted this problem before I read up on it.
    I’m assuming you made the same mistake Chris did on rolling 6 times rather than 5. Otherwise you would have had the answer to c correct, but only by accident.
    I’m not sure exactly how to put in words why we need that first term, so maybe someone else can help me out with that.

  7. 7. Chris Said:

    Hi DP. I hadn’t realised it was the binomial distribution until a while after posting. But for two 3′a I simply realised that I would get the result if I threw 33xxx,3×3xx,3xx3x,etc. That C(5,2) ways altogether. It’s only a formailty to go on to define the binomial distribution from there.

    To get e.g. the “≈” symbol, I originally used charmap. But I keep a bunch of such symbols in a notepad file. The one I just used, I copy/pasted from my post above.

  8. 8. B.tulsi rao Said:

    i think the ans is-[4] 6’s

    but i’m not fully sure about it……..

    becoz out of five chances the probablity of getting a fair no is 4 and out of 6 different no’s……..

  9. 9. DP Said:

    Thanks for the help Chris.

    rao…I’m not sure how to respond to you. I don’t see an answer, so I can’t tell you you are wrong.
    I think you may have thought it was a multiple choice question. It is a multiple part question, which you must answer each part (a, b, & c) with a probability.
    Your last sentence fragment isn’t very clear to me. What are you refering to when you say “fair no”?

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