定量法第一章习题解答 习题2 (1)P{恰有一件次品} 6194 0.0855 200 (2)P{全是正品} 194 0.9122 200 (3)P{至少2件正品}=P{2件正品}+P3件正品} P{1件次品}P{全是正品} =0.0855+0.9122=0.9978
定量法第一章习题解答 习题2． (1) P{恰有一件次品} = =0.0855 (2) P{全是正品}= =0.9122 (3) P{至少2件正品}= P{2件正品}+ P{3件正品} = P{1件次品}+ P{全是正品} =0.0855+0.9122=0.9978 3 200 2 194 1 6 C C C 3 200 3 194 C C
习题4解答 设A={寿命≥50},B={寿命≥70}, 由题意,P(A=1-0.1=0.9, P(B)=1-0.75=0.25 求P(BA) BCA,∴P(AB)P(B) P(BA=P(AB)/P(A)=PB)/P(A 0.25/0.9=0.278
习题4解答 设A={寿命50}，B={寿命70}， 由题意，P(A)=1-0.1=0.9， P(B)=1-0.75=0.25 求P(B|A)。 BA，∴P(AB)=P(B) P(B|A)=P(AB) / P(A)=P(B) / P(A) =0.25/0.9=0.278
习题6解答 设A1、A2、A3分别为抽到用甲、乙、丙厂原料生 产的产品,B={抽到次品},则 P(A1)=0.6,P(A2)=0.3,P(A3)=0.1 P(B|A1)=0.02,P(BA2)=0.03,P(B|A3)=0.05 (1)P(BA3)=P(A3)P(BA3)=0.1×0.05=0.005 (2)P(B)=P(A1)P(BA1)+P(A2)P(BA2) +P(A3)P(B|A3) =0.6×0.02+0.3×003+0.1×0.05=0.026 (3)P(A1B)=P(A1)P(BA1)/P(B) =0.6×0.02/0.026=04615
习题6解答 设A1、A2、A3分别为抽到用甲、乙、丙厂原料生 产的产品，B={抽到次品}，则 P(A1 )=0.6，P(A2 )=0.3，P(A3 )=0.1 P(B|A1 )=0.02，P(B|A2 )=0.03，P(B|A3 )=0.05 (1) P(BA3 )=P(A3 )P(B|A3 )=0.1×0.05=0.005 (2) P(B)=P(A1 )P(B|A1 )+ P(A2 )P(B|A2 ) + P(A3 )P(B|A3 ) =0.6×0.02+0.3×0.03+0.1×0.05=0.026 (3) P(A1 |B)=P(A1 )P(B|A1 ) / P(B) =0.6×0.02 / 0.026=0.4615
习题7解答 设至少应配备n门高炮,并设X为击中敌机的高炮 数,则Ⅹ~B(n,O.02),由题意,使 P(X≥1)=1-P(X=0 =1-C0.02098″=1-098″≥0.3 得0.98007 nlg0.98-g0.7 n≥ 0.7 =1765 lg0.98 故至少应配备18门高炮
习题7解答 设至少应配备n门高炮，并设X为击中敌机的高炮 数，则X～B(n, 0.02)，由题意，使 P(X≥1)=1-P(X=0) 得 0.98n≤0.7 n lg0.98≤lg0.7 故至少应配备18门高炮。 17.65 lg 0.98 lg 0.7 n = 1 0.02 0.98 1 0.98 0.3 0 0 = − = − n n Cn
习题8解答 (1)设X={一合中的次品数},则X~B(n,p), n=100,p=0.01,q=0.99 P{X=0}=C10×0010×099100366 (2)P{X>2}=1-P{X=0}-P{X=1}-P{X=2} 1-0.366-C10X×001×0.99C70×0.012×0998 =0.0794
习题8解答 (1) 设X={一合中的次品数}，则X～B(n，p)， n=100，p=0.01，q=0.99 P{X=0}= ×0.010×0.99100=0.366 (2) P{X>2}=1-P{X=0}-P{X=1}-P{X=2} =1-0.366- ×0.01×0.9999 - ×0.012×0.9998 =0.0794 0 C100 1 C100 2 C100