Question

(a) What is the probability that a 5-card poker hand has at least three spades? (b) What upper bound does Markov's Theorem give for this probability? (c) What upper bound does Chebyshev's Theorem give for this probabil- ity?

Answer 1

Answer:-

Given that:-

Let X be the number of spades in the spades

Total cards$=\binom{52}{5}$

(0 cards from spades & all 5 from there )

) P(X=0) - (0

$P(X=1)=\frac{\binom{13}{1}\binom{39}{4}}{\binom{52}{5}}$

Ci5-3 P(X = i) =

So, this is the card of a hypergeometric distribution

CM-R) P(X = k) = h(k;N, M, n) = { K = 0,1,2,....min(n,m) , otherwise

So, one card N=52 , n= 5, M = 13

(a) What is the probability that a 5-card poker hand has at least three

spades?

$P(X\geq 3)=1-P(X\leq 2)$

$=1-(P(X=0)+P(X=1)+P(X=2))$

  $=1-\frac{\binom{39}{5}}{\binom{52}{5}}-\frac{\binom{13}{1}\binom{39}{5}}{\binom{52}{5}}-\frac{\binom{13}{2}\binom{39}{5}}{\binom{52}{5}}$
(b) What upper bound does Markov’s Theorem give for this probability?

Markov's inequality

If x is a Hypageometric random variable and a>0,

then, $P(X\geq a)\leq \frac{E(X)}{a}$

So, as  

X Hypageometric(52,13,5)

$E(X) = \frac{nM}{N}=\frac{13\times 5}{52}=\frac{65}{52}$

So,

$P(X\geq 3)\leq \frac{65/52}{3}$

$=\frac{65}{156}\simeq 0.416667$

(c) What upper bound does Chebyshev’s Theorem give for this probability?

Chebyshev’s inequaility:-

$P(|X-\mu|\geq \varepsilon )\leq \frac{\sigma ^{2}}{\varepsilon ^{2}}$

$\mu = E(X)$ &  $\sigma ^{2}=V(X)$

So,

$\mu = E(X)= \frac{nM}{N}=\frac{65}{2}$

$\sigma ^{2} = V(X)=n\frac{M}{N}\frac{(N-M)}{N}\frac{N-n}{N-1}$

$=5\frac{13}{52}\, \, \frac{39}{52}\, \, \frac{47}{51}$

$=\frac{5}{4}\times \frac{3}{4}\times \frac{47}{51}$

$=\frac{5\times 47}{16\times 17}$

$=\frac{235}{272}$

So,

$P(X\geq 3)=P(X-\mu\geq 3-\frac{65}{52})$

$=P(X-\mu\geq \frac{91}{52})$

$P(|X-\mu|\geq \varepsilon )=P(X-\mu\geq \varepsilon )+P(X-\mu\leq -\varepsilon )$

$\Rightarrow P(X-\mu\geq \varepsilon )=P(|X-\mu|\geq \varepsilon )+P(X-\mu\leq -\varepsilon )$

$\leq \frac{\sigma ^{2}}{\varepsilon ^{2}}+P(X\leq \mu-\varepsilon )$

$[\mu = \frac{65}{52}\, \, \, \varepsilon =\frac{92}{92}\Rightarrow \mu -\varepsilon =\frac{65}{52}-\frac{91}{52}=-\frac{26}{52}]$

  $\leq \frac{\sigma ^{2}}{\varepsilon ^{2}}+P(X\leq -\frac{26}{52})$ (because X is a positive R-V)

$\leq \frac{\frac{235}{272}}{(\frac{91}{52})^{2}}$

$=0.2821128$