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The first exercises

1. Counting theory:
We may all remember: Given n objects, the number of ways of ordering these objects is:
n! = n(n-1)(n-2)..3.2.1.

n choose k, which is the number of distinct ways of choosing k objects from n: (n k) = n! / ( k! * (n-k)! )

For example:
If we want to form a random group of 3 students among 20 students, there are:
20! / (3! * 17!) = 1140 possible groups

2. Probability
- First problem: Discrete probability distribution for the sum of two dice.


- Second problem: Two people take turns trying to sink a basketball into a net. Person 1 succeeds with probability 1/3, person 2 with 1/4. What is the probability that person 1 succeeds before person 2?
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Comments

  1. Dieu-Thu LeFebruary 9, 2012 at 6:41 PM

    Solution for problem 2:

    Let E be the event that person 1 succeeds before person 2.

    Aj be the event that person 1 succeeds before person 2 and the first success is on trial number j.

    Hence, P(E) = P(A1) + P(A2) + .. + P(A-infinite)

    P(A1): person 1 succeeds on the first trial = 1/3
    P(A2): person 1 misses, 2 misses, then 1 succeeds = 2/3 * 3/4 * 1/3 = 1/2 * 1/3
    .
    .
    .
    P(Aj) = (1/2)^(j-1) * (1/3)

    Therefore,
    P(E) = 2/3

    Note:
    We used: sum_(j=k)^{infinite}{r^j} = r^k/(1-r)

    ReplyDelete

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