NTHU MATH 2810 Midterm Examination Solution Nov 22, 2016 A1, B1 24pts a False False c True d True e False f False g True h True a False False c True d True e False f False g True h True A2, B2 14pts a 3pts 4pts P B E 1 P E 1 BP B P E 1 BP B P E 1 B c P B c 1/8 06 1/8 06 1 04 0075 0475 01579 where P E 2 E 1 P E 1 E 2 P E 1 1/16 06 1/8 06 1 04 00375 0475 00789, P E 1 P E 1 BP B P E 1 B c P B c and 1/8 06 1 04 0475, P E 1 E 2 P E 1 E 2 BP B P E 1 E 2 B c P B c 1/16 06 0 04 00375 An alternative method to get the answer is: P E 2 E 1 P B E 1 P E 2 E 1 B P B c E 1 P E 2 E 1 B c P B E 1 P E 2 B P B c E 1 P E 2 B c I 01579 1/2 1 01579 0 00789 a 3pts 4pts P B E 1 P E 1 BP B P E 1 BP B P E 1 B c P B c 1/8 03 1/8 03 1 07 00375 07375 00508 where P E 2 E 1 P E 1 E 2 P E 1 1/16 03 1/8 03 1 07 001875 07375 00254, P E 1 P E 1 BP B P E 1 B c P B c and 1/8 03 1 07 07375, P E 1 E 2 P E 1 E 2 BP B P E 1 E 2 B c P B c 1/16 03 0 07 001875 An alternative method to get the answer is: P E 2 E 1 P B E 1 P E 2 E 1 B P B c E 1 P E 2 E 1 B c P B E 1 P E 2 B P B c E 1 P E 2 B c II 00508 1/2 1 00508 0 00254 1
A2, B2 cont c 4pts Because P E 2 P E 2 BP B P E 2 B c P B c 1/2 06 0 04 03 P E 2 E 1, the events E 1 and E 2 are not independent d 3pts Yes We used P E 1 E 2 B 1/16 1/8 1/2 P E 1 B P E 2 B and P E 1 E 2 B c 0 1 0 P E 1 B c P E 2 B c in the calculation We used P E 2 E 1 B P E 2 B and P E 2 E 1 B c P E 2 B c to derive I, which also come from the conditional independence assumption c 4pts Because P E 2 P E 2 BP B P E 2 B c P B c 1/2 03 0 07 015 P E 2 E 1, the events E 1 and E 2 are not independent d 3pts Yes We used P E 1 E 2 B 1/16 1/8 1/2 P E 1 B P E 2 B and P E 1 E 2 B c 0 1 0 P E 1 B c P E 2 B c in the calculation We used P E 2 E 1 B P E 2 B and P E 2 E 1 B c P E 2 B c to derive II, which also come from the conditional independence assumption A3, B3 14pts a 4pts Because X Poisson p P X 8 λ x8 400, 000 100, 000 4, e 4 4 x 1 7 e 4 4 x x0 4pts Because Y inomial12, p, 12 12 P Y 2 p y 12 y y y2 1 12 y 12 y y y0 c 4pts Because Z geometricp, P Z 6 1 z 1 p z6 5 z 1 p z1 a 4pts Because X Poisson p P X 5 λ x5 400, 000 50, 000 8, e 8 8 x 1 4 e 8 8 x x0 4pts Because Y geometricp, P Y 7 1 y 1 p y7 6 y 1 p y1 c 4pts Because Z inomial12, p, 12 12 P Z 4 z z4 3 12 1 z z0 p z 12 z p z 12 z 2
A3, B3 cont d 2pts We are making the assumptions of independence: i one person s act of suicide does not influence another person to commit or not suicide, and ii different Bernoulli trials at least 8 suicides or not in every months are independent d 2pts We are making the assumptions of independence: i one person s act of suicide does not influence another person to commit or not suicide, and ii different Bernoulli trials at least 5 suicides or not in every months are independent A4, B4 18pts a 3pts Because P N 0 1 1 n1 P N n n1 αp n 1 αp, the proaility mass function of N is 1 αp, if n 0, f N n αp n, if n 1, 2, 3,, 0, otherwise 4pts Denote the cumulative distriution function of N y F N n For k n < k 1, k 1, 2, 3,, F N n 1 αp k αp i i1 1 αp αp αpk1 1 αpk1 So, 0, if n < 0, 1 αp, if 0 n < 1, F N n 1 αpk1, if k n < k 1, k 1, 2, 3, c 3pts Because the distriution of B given N n is inomialn, 1/2, n P B N n 1/2 n a 3pts Because P N 0 1 1 n1 P N n n1 αp n1 1 αp2, the proaility mass function of N is 1 αp2, if n 0, f N n αp n1, if n 1, 2, 3,, 0, otherwise 4pts Denote the cumulative distriution function of N y F N n For k n < k 1, k 1, 2, 3,, F N n 1 αp2 k i1 1 αp2 αp2 αp k2 So, αp i1 1 αpk2 0, if n < 0, 1 αp2, if 0 n < 1, F N n 1 αpk2, if k n < k 1, k 1, 2, 3, c 3pts Because the distriution of B given N n is inomialn, 1/2, n P B N n 1/2 n 3
A4, B4 cont d 4pts The question asks to find P B By the law of total proaility, for 0, P B 1 P B N np N n n0 1 αp n1 1 αp α p/2 n, n1 and for 1, 2, 3,, P B n 1/2 n αp n 0 P B N np N n n n α n 1/2 n αp n p/2 n e 4pts The question asks to find P N n B By the Bayes rule, P N n B For 0, P B N np N n P B P N n B 1 αp 1 αp α, if n 0, i1 p/2i αp/2 n, if n 1, 2, 3,, i1 p/2i 1 αp α and for 1, 2, 3,, n, 1, 2,, P N n B p/2 n i i p/2 i d 4pts The question asks to find P B By the law of total proaility, for 0, P B 1 P B N np N n n0 1 αp2 n1 1 αp2 αp p/2 n, n1 and for 1, 2, 3,, P B n 1/2 n αp n1 0 P B N np N n n n αp n 1/2 n αp n1 n p/2 n e 4pts The question asks to find P N n B By the Bayes rule, P N n B For 0, P N n B 1 αp2 1 αp2 αp i1 αpp/2 n 1 αp2 αp i1 P B N np N n P B p/2i, if n 0, p/2i, if n 1, 2, 3,, and for 1, 2, 3,, n, 1, 2,, p/2 n P N n B i i p/2 i A5, B5 19pts a 2pts Unlike the X r is to withdraw without replacement, in the negative inomial, the all withdrawn is replaced efore the next drawing The negative inomial is constructed from a sequence of independent Bernoulli trials, while X r is constructed from dependent Bernoulli trials 4
5pts Let A x e the event that a red all is otained on the xth draw Let Z x 1 e the numer of red alls withdrawn on the first x 1 draws Then, Z x 1 hypergeometricx 1, N, R, and P X r x P {Z x 1 r 1} A x P Z x 1 r 1P A x Z x 1 r 1 Because P A x Z x 1 r 1 R r 1, R N R p r x P X r x r 1 x r N R r 1, x r, r 1,, N, x 1 where s t 0 if s < t c 5pts EX r xp r x xr Because Y r X r r, d 7pts Because and N R x r R! r 1!R r 1! 1 xr x N! x 1!!! N R r!r r 1! R r 1 N 1r x r R r 1 N 1! xr! N 1 N 1r r 1 1 x 1 r 1 r 1 1 xr N 1 N 1 x 1 1 x 1 1 N 1 N1 N 1r r 1 1 y r 1 r 1 1 yr1 N 1 N 1 y 1 y 1 }{{} pmf of negative hypergeometricr 1, N 1, Note let y e x 1 N 1r 1 r N 1 N 1 EY r EX r r r 1 r N R V arx r EX 2 r [EX r ] 2 EX 2 r EX r EX r [EX r ] 2 E[X r X r 1] EX r [EX r ] 2, 5
E[X r X r 1] xx 1p r x xr N 1N 2rr 1 R 2 R! N R r 1!R r 1! x r 1 xr x 1x N! x 1!! R 2! r 1!R r 1! N R x r R r 1 R r 1 N 2! xr x 1!! R 2 N 2 R 2 N 1N 2rr 1 r 2 1 x 2 r 2 R 2 r 2 1 R 2 xr N 2 N 2 x 2 1 x 2 1 R 2 N 2 R 2 N2 N 1N 2rr 1 r 2 1 y r 2 R 2 r 2 1 R 2 yr2 N 2 N 2 y 1 y 1 }{{} pmf of negative hypergeometricr 2, N 2, R 2 Note let y e x 2 N 1N 2rr 1, R 2 V arx r N 1N 2rr 1 rn 1 R 2 r2 N 1 2 r 2 N RN 1R r 1 2 R 2 A6, B6 11pts Because adding a constant to a random variale does not change its variance and Y r X r r, V ary r V arx r a 5pts We have E T 1 T 2 T 3 T 4 T 5 Because T 1,, T 5 are independent events, P E P T 1 T 2 T 3 T 4 P T 5 {1 [1 P T 1 T 2 ][1 P T 3 T 4 ]} P T 5 {1 [1 P T 1 P T 2 ][1 P T 3 P T 4 ]} P T 5 [1 1 p 2 3 p 4 ]p 5 p 1 p 2 p 3 p 4 p 1 p 2 p 3 p 4 p 5 a 5pts We have E T 1 T 3 T 2 T 4 T 5 Because T 1,, T 5 are independent events, P E P T 1 T 3 T 2 T 4 P T 5 {1 [1 P T 1 T 3 ][1 P T 2 T 4 ]} P T 5 {1 [1 P T 1 P T 3 ][1 P T 2 P T 4 ]} P T 5 [1 1 p 3 2 p 4 ]p 5 p 1 p 3 p 2 p 4 p 1 p 2 p 3 p 4 p 5 6
A6, B6 cont 6pts When T 3 occurs relay 3 closes, the system works current can flow etween A and B if the event G 1 T 1 T 2 T 4 T 5 also occurs, and ecause T 1,, T 5 are independent events, P G 1 P T 1 T 2 P T 4 T 5 {1 [1 P T 1 ][1 P T 2 ]} {1 [1 P T 4 ][1 P T 5 ]} [1 1 2 ] [1 4 5 ] p 1 p 2 p 1 p 2 p 4 p 5 p 4 p 5 When T c 3 occurs relay 3 opens, the system works if the event G 2 T 1 T 4 T 2 T 5 also occurs, and ecause T 1,, T 5 are independent, P G 2 1 [1 P T 1 T 4 ][1 P T 2 T 5 ] 1 [1 P T 1 P T 4 ][1 P T 2 P T 5 ] 1 1 p 4 2 p 5 p 1 p 4 p 2 p 5 p 1 p 2 p 4 p 5 So, E G 1 T 3 G 2 T c 3 {[T 1 T 2 T 4 T 5 ] T 3 } {[T 1 T 4 T 2 T 5 ] T c 3 }, and note that G 1 T 3 and G 2 T3 c are disjoint ecause T 3 and T3 c form a partition of the sample space Because T 1,, T 5 are independent, P E P G 1 T 3 P G 2 T c 3 P G 1 T 3 P T 3 P G 2 T c 3 P T c 3 P G 1 P T 3 P G 2 P T c 3 p 1 p 2 p 1 p 2 p 4 p 5 p 4 p 5 p 3 p 1 p 4 p 2 p 5 p 1 p 2 p 4 p 5 3 6pts When T 3 occurs relay 3 closes, the system works current can flow etween A and B if the event G 1 T 1 T 4 T 2 T 5 also occurs, and ecause T 1,, T 5 are independent events, P G 1 P T 1 T 4 P T 2 T 5 {1 [1 P T 1 ][1 P T 4 ]} {1 [1 P T 2 ][1 P T 5 ]} [1 1 4 ] [1 2 5 ] p 1 p 4 p 1 p 4 p 2 p 5 p 2 p 5 When T c 3 occurs relay 3 opens, the system works if the event G 2 T 1 T 2 T 4 T 5 also occurs, and ecause T 1,, T 5 are independent, P G 2 1 [1 P T 1 T 2 ][1 P T 4 T 5 ] 1 [1 P T 1 P T 2 ][1 P T 4 P T 5 ] 1 1 p 2 4 p 5 p 1 p 2 p 4 p 5 p 1 p 2 p 4 p 5 So, E G 1 T 3 G 2 T c 3 {[T 1 T 4 T 2 T 5 ] T 3 } {[T 1 T 2 T 4 T 5 ] T c 3 }, and note that G 1 T 3 and G 2 T3 c are disjoint ecause T 3 and T3 c form a partition of the sample space Because T 1,, T 5 are independent, P E P G 1 T 3 P G 2 T c 3 P G 1 T 3 P T 3 P G 2 T c 3 P T c 3 P G 1 P T 3 P G 2 P T c 3 p 1 p 4 p 1 p 4 p 2 p 5 p 2 p 5 p 3 p 1 p 2 p 4 p 5 p 1 p 2 p 4 p 5 3 7