CS206 -Discrete Mathematics II
Instructor: Chitoor V.Srinivasan
PROBLEM SET 9 SOLUTIONS

  1. Let n be the number of random digits and let X be the number of occurrences of the digit 7. We want tex2html_wrap_inline101 . Solving for n yields   tex2html_wrap_inline103 ,  hence  n = 22.

  2. Let's consider the people to be indistinguishable (considering to be distinct yields the same answer but much more difficultly). There are   tex2html_wrap_inline107   ways to partition the 6 birthdays into 12 months. To find the number which fall into 2 calendar months let's choose 2 months which can be done in   tex2html_wrap_inline109   ways. For each such choice of months, their are   tex2html_wrap_inline111   non-empty partitions of 6 birthdays into the two months. Thus the probability is

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  3. It's a Binomial distribution.   tex2html_wrap_inline113 . Solving for   tex2html_wrap_inline115   we get   tex2html_wrap_inline117   and   tex2html_wrap_inline119 ,  which is approximately zero. Note: some biologists claim that the chances of a simple bacterium forming from the random collisions of atoms and molecules is even much more remote than that!

  4. It's Poisson distribution, with   tex2html_wrap_inline121 .

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  5. It's Bernoulli trial, with   tex2html_wrap_inline123 .
    (a).
    It's Negative binomial distribution with r = 5 successes.

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    (b).
    It's also Negative binomial distribution with r = 5 successes.

    eqnarray43

    (c).
    It's Binomial distribution, with  n = 7.

    eqnarray46

  6. This requires use of negative binomial distribution in which ``success'' is producing a defective item. The answer is the probability that five or more tries are made before stopping, where the stopping condition is, obtaining 3 successes (i.e. getting 3 defective items). This is,

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    The expected number of trials for the negative binomial distribution is 3/(.15) = 20.

  7. Lets define random variable  Head  to be the no.of heads, and  Tail  to be the no.of tails. These are both binomial distributions so   tex2html_wrap_inline131   and   tex2html_wrap_inline133 . Thus,

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  8. When   tex2html_wrap_inline135   and   tex2html_wrap_inline137 .

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    So if the ticket is $5, it's better to bet, and if it's $10, it's better not to bet.

  9. Lets define the random value   tex2html_wrap_inline139   to be the result of the dice rolled first, lets define the random value   tex2html_wrap_inline141   to be the result of the dice rolled second. We have
    (a).

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    (b).

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    (c).
    Lets assume the dice rolled first getting  i  and dice rolled second getting  j,  and let   tex2html_wrap_inline147   correspond to the entrance  (i,j)  of the following.

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    eqnarray65

    (d).
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    (e).
    tex2html_wrap_inline153   5s and 6s on 1st die)   tex2html_wrap_inline155   5s on 1st die)   tex2html_wrap_inline157   6s on 1st die)]   = 2 (1/6 + 1/6) = 2/3