Raccolta delle soluzioni essenziali degli esercizi di approfondimento (dai Miro 4G), raggruppate per tema: le tre rappresentazioni, la combinatoria pura, lo schema binomiale e le scommesse. Per gli enunciati completi si vedano i singoli esercizi svolti del capitolo.
Tre rappresentazioni
P ( B ) = 0 , 4 ⋅ 1 + 0 , 6 ⋅ 0 , 2 = 0 , 52 P(B)=0,4\cdot 1 + 0,6\cdot 0,2 = 0,52 P ( B ) = 0 , 4 ⋅ 1 + 0 , 6 ⋅ 0 , 2 = 0 , 52 P ( A ∣ B ) = 0 , 4 / 0 , 52 = 10 / 13 ≈ 0 , 77 P(A\mid B)=0,4/0,52=10/13\approx 0,77 P ( A ∣ B ) = 0 , 4/0 , 52 = 10/13 ≈ 0 , 77 P ( ? , ? , V ) = 5 8 ⋅ 4 7 ⋅ 3 6 + 5 8 ⋅ 3 7 ⋅ 4 6 + 3 8 ⋅ 5 7 ⋅ 4 6 + 3 8 ⋅ 2 7 ⋅ 5 6 = 5 8 P(?,?,V)=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{3}{8}\cdot\frac{5}{7}\cdot\frac{4}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{5}{6}=\frac{5}{8} P ( ? , ? , V ) = 8 5 ⋅ 7 4 ⋅ 6 3 + 8 5 ⋅ 7 3 ⋅ 6 4 + 8 3 ⋅ 7 5 ⋅ 6 4 + 8 3 ⋅ 7 2 ⋅ 6 5 = 8 5 P ( V V V ) = 5 8 ⋅ 4 7 ⋅ 3 6 = 5 28 P(VVV)=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{5}{28} P ( V V V ) = 8 5 ⋅ 7 4 ⋅ 6 3 = 28 5 5 / 28 5 / 8 = 8 28 = 2 7 \frac{5/28}{5/8}=\frac{8}{28}=\frac{2}{7} 5/8 5/28 = 28 8 = 7 2 Tabella: vaccinati ammalati = 6000 ⋅ 0 , 1 = 600 =6000\cdot 0,1=600 = 6000 ⋅ 0 , 1 = 600 = 4000 ⋅ 0 , 3 = 1200 =4000\cdot 0,3=1200 = 4000 ⋅ 0 , 3 = 1200 = 1800 =1800 = 1800 P ( V ‾ ∣ ammalato ) = 1200 / 1800 = 2 / 3 P(\overline{V}\mid \text{ammalato})=1200/1800=2/3 P ( V ∣ ammalato ) = 1200/1800 = 2/3
Combinatoria
26 ! 2 ! ⋅ 1 ! ⋅ 5 ! ⋅ 1 ! ⋅ 4 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 1 ! \dfrac{26!}{2!\cdot 1!\cdot 5!\cdot 1!\cdot 4!\cdot 2!\cdot 2!\cdot 2!\cdot 2!\cdot 2!\cdot 2!\cdot 1!} 2 ! ⋅ 1 ! ⋅ 5 ! ⋅ 1 ! ⋅ 4 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 2 ! ⋅ 1 ! 26 ! ( 20 + 3 − 1 3 ) = ( 22 3 ) = 1540 \binom{20+3-1}{3}=\binom{22}{3}=1540 ( 3 20 + 3 − 1 ) = ( 3 22 ) = 1540 26 4 = 456 976 26^4=456\,976 2 6 4 = 456 976 (a) 6 ! = 720 6!=720 6 ! = 720 5 ! ⋅ 2 ! = 240 5!\cdot 2!=240 5 ! ⋅ 2 ! = 240 4 ! ⋅ 2 ! ⋅ 2 = 96 4!\cdot 2!\cdot 2=96 4 ! ⋅ 2 ! ⋅ 2 = 96
(a) ( 28 5 ) = 98 280 \binom{28}{5}=98\,280 ( 5 28 ) = 98 280 ( 4 2 ) ( 28 3 ) = 6 ⋅ 3276 = 19 656 \binom{4}{2}\binom{28}{3}=6\cdot 3276=19\,656 ( 2 4 ) ( 3 28 ) = 6 ⋅ 3276 = 19 656 ( 8 2 ) ( 8 3 ) = 28 ⋅ 56 = 1568 \binom{8}{2}\binom{8}{3}=28\cdot 56=1568 ( 2 8 ) ( 3 8 ) = 28 ⋅ 56 = 1568 ( 4 3 ) ( 28 2 ) + ( 4 4 ) ( 28 1 ) = 4 ⋅ 378 + 28 = 1540 \binom{4}{3}\binom{28}{2}+\binom{4}{4}\binom{28}{1}=4\cdot 378 + 28=1540 ( 3 4 ) ( 2 28 ) + ( 4 4 ) ( 1 28 ) = 4 ⋅ 378 + 28 = 1540
Schema binomiale
( 6 2 ) ( 1 / 6 ) 2 ( 5 / 6 ) 4 = 15 ⋅ ( 1 / 36 ) ⋅ ( 625 / 1296 ) ≈ 0 , 2 \binom{6}{2}(1/6)^2(5/6)^4=15\cdot (1/36)\cdot (625/1296)\approx 0,2 ( 2 6 ) ( 1/6 ) 2 ( 5/6 ) 4 = 15 ⋅ ( 1/36 ) ⋅ ( 625/1296 ) ≈ 0 , 2 1 − ( 0 , 3 ) 4 = 1 − 0 , 0081 = 0 , 9919 1-(0,3)^4 = 1-0,0081 = 0,9919 1 − ( 0 , 3 ) 4 = 1 − 0 , 0081 = 0 , 9919 P 50 ( 1 ) = 50 ⋅ 0 , 02 ⋅ ( 0 , 98 ) 49 ≈ 0 , 371 P_{50}(1)=50\cdot 0,02\cdot (0,98)^{49}\approx 0,371 P 50 ( 1 ) = 50 ⋅ 0 , 02 ⋅ ( 0 , 98 ) 49 ≈ 0 , 371 P ( k ≥ 1 ) = 1 − ( 0 , 98 ) 50 ≈ 0 , 636 P(k\ge 1)=1-(0,98)^{50}\approx 0,636 P ( k ≥ 1 ) = 1 − ( 0 , 98 ) 50 ≈ 0 , 636
Scommesse
(a) vincita netta = 1 , 5 ⋅ 50 = 75 =1,5\cdot 50=75 = 1 , 5 ⋅ 50 = 75 V ˉ = 0 , 3 ⋅ 75 + 0 , 7 ⋅ ( − 50 ) = 22 , 5 − 35 = − 12 , 5 \bar V=0,3\cdot 75 + 0,7\cdot(-50)=22,5-35=-12,5 V ˉ = 0 , 3 ⋅ 75 + 0 , 7 ⋅ ( − 50 ) = 22 , 5 − 35 = − 12 , 5 0 , 3 ⋅ x + 0 , 7 ⋅ ( − 50 ) = 0 ⇒ x = 350 / 3 ≈ 116 , 67 0,3\cdot x + 0,7\cdot(-50)=0 \Rightarrow x=350/3\approx 116,67 0 , 3 ⋅ x + 0 , 7 ⋅ ( − 50 ) = 0 ⇒ x = 350/3 ≈ 116 , 67 ( 50 + 116 , 67 ) / 50 ≈ 3 , 33 : 1 (50+116,67)/50\approx 3,33:1 ( 50 + 116 , 67 ) /50 ≈ 3 , 33 : 1 12 , 5 / 50 = 25 % 12,5/50=25\% 12 , 5/50 = 25%
V ˉ = 1 6 ⋅ 30 + 1 6 ⋅ 6 + 4 6 ⋅ ( − 5 ) = 5 + 1 − 3 , 33 = 2 , 67 \bar V=\frac{1}{6}\cdot 30 + \frac{1}{6}\cdot 6 + \frac{4}{6}\cdot(-5)=5+1-3,33=2,67 V ˉ = 6 1 ⋅ 30 + 6 1 ⋅ 6 + 6 4 ⋅ ( − 5 ) = 5 + 1 − 3 , 33 = 2 , 67 ≈ 2 , 67 \approx 2,67 ≈ 2 , 67
Collegamenti
Argomenti: Probabilita Concetti: Calcolo combinatorio · Distribuzione binomiale · Gioco equo · Teorema di bayes · Valore atteso