This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Kuna shanga 5 nyekundu, 4 nyeupe, na 3 za bluu. Jumla ya shanga ni 5 + 4 + 3 = 12. Shanga mbili huchaguliwa bila kurudishwa. a) Chora mchoro wa mti na uorodheshe nafasi ya uwezekano. Uchaguzi wa kwanza: Uwezekano wa shanga nyekundu (R): (5)/(12) Uwezekano wa shanga nyeupe (W): (4)/(12) Uwezekano wa shanga ya bluu (B): (3)/(12) Uchaguzi wa pili (baada ya shanga moja kuchaguliwa): Ikiwa shanga ya kwanza ilikuwa nyekundu (R1): Zimebaki shanga 11 (4R, 4W, 3B). P(R2|R1) = (4)/(11) P(W2|R1) = (4)/(11) P(B2|R1) = (3)/(11) Ikiwa shanga ya kwanza ilikuwa nyeupe (W1): Zimebaki shanga 11 (5R, 3W, 3B). P(R2|W1) = (5)/(11) P(W2|W1) = (3)/(11) P(B2|W1) = (3)/(11) Ikiwa shanga ya kwanza ilikuwa ya bluu (B1): Zimebaki shanga 11 (5R, 4W, 2B). P(R2|B1) = (5)/(11) P(W2|B1) = (4)/(11) P(B2|B1) = (2)/(11) Mchoro wa mti ungekuwa na matawi matatu kwa uchaguzi wa kwanza (R, W, B), na kila tawi likigawanyika katika matawi matatu zaidi kwa uchaguzi wa pili (R, W, B), na kuonyesha uwezekano kwenye kila tawi. Nafasi ya uwezekano (sample space) na uwezekano wake ni: P(R_1, R_2) &= (5)/(12) × (4)/(11) = (20)/(132) \\ P(R_1, W_2) &= (5)/(12) × (4)/(11) = (20)/(132) \\ P(R_1, B_2) &= (5)/(12) × (3)/(11) = (15)/(132) \\ P(W_1, R_2) &= (4)/(12) × (5)/(11) = (20)/(132) \\ P(W_1, W_2) &= (4)/(12) × (3)/(11) = (12)/(132) \\ P(W_1, B_2) &= (4)/(12) × (3)/(11) = (12)/(132) \\ P(B_1, R_2) &= (3)/(12) × (5)/(11) = (15)/(132) \\ P(B_1, W_2) &= (3)/(12) × (4)/(11) = (12)/(132) \\ P(B_1, B_2) &= (3)/(12) × (2)/(11) = (6)/(132) b) Tafuta uwezekano kwamba i) Shanga ya mwisho iliyochaguliwa ni nyekundu. Hii inamaanisha shanga ya pili ni nyekundu. Tunajumlisha uwezekano wa matokeo ambapo shanga ya pili ni nyekundu: (R_1, R_2), (W_1, R_2), (B_1, R_2). P(shanga ya mwisho ni nyekundu) = P(R_1, R_2) + P(W_1, R_2) + P(B_1, R_2) = (20)/(132) + (20)/(132) + (15)/(132) = (55)/(132) Kurahisisha sehemu: (55)/(132) = (55 ÷ 11)/(132 ÷ 11) = (5)/(12) ii) Shanga zilizochaguliwa zilikuwa za rangi moja. Hii inamaanisha shanga zote mbili ni nyekundu, au zote nyeupe, au zote za bluu. Tunajumlisha uwezekano wa matokeo: (R_1, R_2), (W_1, W_2), (B_1, B_2). P(rangi moja) = P(R_1, R_2) + P(W_1, W_2) + P(B_1, B_2) = (20)/(132) + (12)/(132) + (6)/(132) = (38)/(132) Kurahisisha sehemu: (38)/(132) = (38 ÷ 2)/(132 ÷ 2) = (19)/(66) iii) Angalau moja ya shanga zilizochaguliwa ni ya bluu. Hii inaweza kuhesabiwa kama 1 - P(hakuna shanga ya bluu). Idadi ya shanga zisizo za bluu = 5 (nyekundu) + 4 (nyepe) = 9. P(hakuna shanga ya bluu) = P(ya kwanza si bluu) × P(ya pili si bluu | ya kwanza si bluu) = (9)/(12) × (8)/(11) = (72)/(132) Kurahisisha sehemu: (72)/(132) = (72 ÷ 12)/(132 ÷ 12) = (6)/(11) Sasa, P(angalau moja ni bluu) = 1 - P(hakuna shanga ya bluu) = 1 - (6)/(11) = (11)/(11) - (6)/(11) = (5)/(11) Tuma swali linalofuata 📸