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 probability?

Answer 1

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Aata from the given question cu 3 clobe- (31739) - 211926 u clubs. 1938134) = 27995 s clubs - (13) ; 1281 probability = 211926+27885 + 1287 - 241098. © This upper bound for the probability that non- negative fonction of a random variable is greater than or equal to some positive constant. It 99 named oflar the Russion mathematical Anotrey markor, although t appeared earlier in the work, If 298 a non-negative Random variable caro), Then the probability that is at least a is at most the espectation of x divided by a : P[220] < 02 Let a= a, E(X). [where ã >0 ) Then we can rewrite the previous inequality as plusã, EL»)) - 5 MEYEXIf (r)>E}< f ldp, mono to nically Inoreasing :- P(1X129) s ECY (1x1)) 420)

PCIYLA) < 6(1X1") an Cheby shew's inequality: X:N y R 13 any Random variable, and let, rso be any positive red no P118-6(1)(20) L anon * tiyst proof of this theorem; Let AGM be defined by : A -{5€ 111X65) - Elx]23)} V(2) = PS)[xc5) – 671) P(s) (265) – E (732) + sp(57 [17)- E(77%) » £ PCS) (765)– El71)} (since vs, Cxcs) - E(71)20. SEA 2 £ P(s) 84( 623 1365)- E(X)] 27 for all SEA. SEN SEA - 098 PLS) - opla) SEA Thus we wanted V(x) > ? P(A). In other words is that to prove. * second prool: A = { senll X(5) - E(x) >>}

= { sen 1 (06)-1(13) 207] 465) = (x (5) -- 1(0))? Pln) - PLY>>') < U(9) . [(x-1(2))") PIA): v(n)