Expected value, variance and hypergeometric distribution

conditional probabilityexpected valueprobabilityvariance

Marco plays the following simplified lotto game: from an urn
containing 10 balls numbered from 1 to 10, 4 balls are extracted without
reinsertion. Marco bets on the pair of numbers {5, 10}. Let X be the
random variable that counts how many numbers Marco guesses, i.e. how
many of the numbers {5, 10} are extracted.

a) Identify the type of distribution of X, and calculate its expected
value and variance.

My attempt:

I think that I can use a hypergeometric random variable (because, among other things, it says that the extraction process is without reinsertion).

$P(X=k) = \frac{\binom{m}{k}\binom{n-m}{r-k}}{{\binom{n}{r}}}$ where n is the total number of elements that we have (in this case, n==10), m is equal to 2 (5 and 10) and r is equal to the number of extracted elements (in this case, r==4).

$P(X=0) = \frac{\binom{2}{0}\binom{10-2}{4-0}}{{\binom{10}{4}}} = \frac{\binom{2}{0}\binom{8}{4}}{{\binom{10}{4}}} = \frac{1}{3}$

$P(X=1) = \frac{\binom{2}{1}\binom{10-2}{4-1}}{{\binom{10}{4}}} = \frac{\binom{2}{1}\binom{8}{3}}{{\binom{10}{4}}} = \frac{8}{15}$

$P(X=2) = \frac{\binom{2}{2}\binom{10-2}{4-2}}{{\binom{10}{4}}} = \frac{\binom{2}{2}\binom{8}{2}}{{\binom{10}{4}}} = \frac{2}{15}$

  • $E(X) = 0p(0) + 1p(1) + 2p(2) = 0P(X=0) + 1P(X=1) + 2P(X=2) = 0\frac{1}{3} + 1\frac{8}{15} + 2\frac{2}{15} = \frac{4}{5}$

Another easier method is, as someone suggested in the comments, remember that if X is an hypergeometric function, $E(X)=r\frac{m}{n}$

$\binom{n}{k}=\frac{n!}{(k!)(n-k!)}$

$\binom{n}{k}=\frac{n}{k}\frac{(n-1)!}{(k-1)!)(n-1-(k-1)!)}=\frac{n}{k}\binom{n-1}{k-1}$

$E(X)=\sum_{k=0}^{n} kp(k)$

$E(X)=\sum_{k=0}^{n} kP(X=k)$

$E(X)=\sum_{k=0}^{n} k\frac{\binom{m}{k}\binom{n-m}{r-k}}{{\binom{n}{r}}}$

$E(X)=\sum_{k=1}^{n} k\frac{\binom{m}{k}\binom{n-m}{r-k}}{{\binom{n}{r}}}$ since for $x=0$ the addend is 0.

$E(X)=\sum_{k=1}^{n} k\frac{\frac{m}{k}\binom{m-1}{k-1}\binom{n-1-(m-1)}{r-1-(k-1)}}{{\frac{n}{r}\binom{n-1}{r-1}}}$

$E(X)=r\frac{m}{n}\sum_{k=1}^{n} \frac{\binom{m-1}{k-1}\binom{n-1-(m-1)}{r-1-(k-1)}}{\binom{n-1}{r-1}}$ and if we say that $a=k-1$

$E(X)=r\frac{m}{n}\sum_{a=0}^{n-1} \frac{\binom{m-1}{a}\binom{n-1-(m-1)}{r-1-(a)}}{\binom{n-1}{r-1}}$

The sum is 1, so $E(X)=r\frac{m}{n} = 4\frac{2}{10} = \frac{4}{5}$.

  • $Var(X) = \frac{rp(1-p)(n-r)}{n-1} = \frac{4\frac{2}{40}(1-\frac{2}{40})(40-4)}{40-1} = \frac{57}{325}$ where $p=\frac{m}{n}=\frac{2}{40}$

Is this correct?

Luca and Andrea also play together with Marco. Luca points to the
triad of numbers {1, 5, 10}, and let Y be the random variable that
counts the numbers guessed by Luca. Instead Andrea flips a
well-balanced coin: if heads came up he bets on the pair {5, 10},
while if tails came up he bets on the triad of numbers {1, 5, 10}. Let
Z be the random variable that counts how many numbers are identified
by Andrea.

b) Write in terms of the events T = {heads} and C = {tails} and the random variables, the following events E, F, G and A, and calculate their probabilities:

  • E = {Marco doesn't even guess a number}
  • F = {Luca doesn't even guess a number}
  • G = {Andrea doesn't even guess a number}
  • A = {Both Marco and Andrea don't guess a single number}

My attempt:

  • E = {Marco doesn't even guess a number} = $(X=0)$

    $P(E) = P(X=0) = \frac{\binom{2}{0}\binom{10-2}{4-0}}{{\binom{10}{4}}} = \frac{\binom{2}{0}\binom{8}{4}}{{\binom{10}{4}}} = \frac{1}{3}$

  • F = {Luca doesn't even guess a number} = $(Y=0)$

    $P(F) = P(Y=0) = \frac{\binom{3}{0}\binom{10-3}{4}}{{\binom{10}{4}}} = \frac{\binom{3}{0}\binom{7}{4}}{{\binom{10}{4}}} = \frac{1}{6}$

  • G = {Andrea doesn't even guess a number} = $(Z=0)$

    • Let's define H={head comes out}
    • Let's define C={cross comes out}

    $P(G) = P(G|H)P(H)+P(G|C)P(C)$
    = $\frac{\binom{2}{0}\binom{10-2}{4-0}}{{\binom{10}{4}}}\frac{1}{2} + \frac{\binom{3}{0}\binom{10-3}{4}}{{\binom{10}{4}}}\frac{1}{2} = \frac{1}{3}\frac{1}{2} + \frac{1}{6}\frac{1}{2} = \frac{1}{4}$

  • A = {Both Marco and Andrea don't guess a single number} = $(X=0)\cap(G=0)$

    $P_{X,Z}(x,z)=P_{Z}(z)P(X=x |Z=z)$

    $P_{X,Z}(0,0)=P_{Z}(0)P(X=0 |Z=0)=P(Z=0)P(X=0 |Z=0)=\frac{1}{4}1=\frac{1}{4}$

Is this correct?

c) Knowing that Andrew didn't guess a single number, calculate the
probability that heads and tails came up.

  • G = {Andrea doesn't even guess a number} = $(Z=0)$

  • Let's define H={head comes out}

  • Let's define C={cross comes out}

    $P(H|G)=\frac{P(G|H)P(H)}{P(G)}=\frac{P(X=0)\frac{1}{2}}{P(Z=0)}=\frac{\frac{1}{3}\frac{1}{2}}{\frac{1}{4}}=\frac{2}{3}$

    $P(C|G)=\frac{P(G|C)P(C)}{P(G)}=\frac{P(Y=0)\frac{1}{2}}{P(Z=0)}=\frac{\frac{1}{6}\frac{1}{2}}{\frac{1}{4}}=\frac{1}{3}$

Is this correct?

Best Answer

In part A, you made some arithmetic mistakes in calculating the binomial coefficients for $P(X=2)$ and $P(X=3)$. In calculating the mean, you forgot to multiply $P(X=2)$ by $2$ (but this calculation uses the incorrect values of $P(X=2)$ and $P(X=3)$. Also note that for any Hypergeometric random variable, the mean is simply $r\tfrac{m}{n}=2(\tfrac{4}{10})=\tfrac45$. It's also safe to assume that you should redo the computation of the variance, though again for common distributions like the Hypergeometric it may be worth memorizing the mean and variance.

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