A bag contains 5 red, 4 white and 3 blue beads. Two beads are selected at random. Draw a tree diagram and list the probability space. Find the probability that: (i) The last bead selected is red. (ii) The beads selected were of the same colour. (iii) At least one of the selected beads is blue.

Mathematics
A bag contains 5 red, 4 white and 3 blue beads. Two beads are selected at random. Draw a tree diagram and list the probability space. Find the probability that: (i) The last bead selected is red. (ii) The beads selected were of the same colour. (iii) At least one of the selected beads is blue.

ScanSolve AI Answer

Kuna shanga 5 nyekundu, 4 nyeupe, na 3 za bluu. Jumla ya shanga ni 5+4+3=125 + 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): 512\frac{5}{12}
    • Uwezekano wa shanga nyeupe (W): 412\frac{4}{12}
    • Uwezekano wa shanga ya bluu (B): 312\frac{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) = 411\frac{4}{11}
      • P(W2|R1) = 411\frac{4}{11}
      • P(B2|R1) = 311\frac{3}{11}
    • Ikiwa shanga ya kwanza ilikuwa nyeupe (W1): Zimebaki shanga 11 (5R, 3W, 3B).
      • P(R2|W1) = 511\frac{5}{11}
      • P(W2|W1) = 311\frac{3}{11}
      • P(B2|W1) = 311\frac{3}{11}
    • Ikiwa shanga ya kwanza ilikuwa ya bluu (B1): Zimebaki shanga 11 (5R, 4W, 2B).
      • P(R2|B1) = 511\frac{5}{11}
      • P(W2|B1) = 411\frac{4}{11}
      • P(B2|B1) = 211\frac{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(R1,R2)=512×411=20132P(R1,W2)=512×411=20132P(R1,B2)=512×311=15132P(W1,R2)=412×511=20132P(W1,W2)=412×311=12132P(W1,B2)=412×311=12132P(B1,R2)=312×511=15132P(B1,W2)=312×411=12132P(B1,B2)=312×211=6132\begin{align*} P(R_1, R_2) &= \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} \\ P(R_1, W_2) &= \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} \\ P(R_1, B_2) &= \frac{5}{12} \times \frac{3}{11} = \frac{15}{132} \\ P(W_1, R_2) &= \frac{4}{12} \times \frac{5}{11} = \frac{20}{132} \\ P(W_1, W_2) &= \frac{4}{12} \times \frac{3}{11} = \frac{12}{132} \\ P(W_1, B_2) &= \frac{4}{12} \times \frac{3}{11} = \frac{12}{132} \\ P(B_1, R_2) &= \frac{3}{12} \times \frac{5}{11} = \frac{15}{132} \\ P(B_1, W_2) &= \frac{3}{12} \times \frac{4}{11} = \frac{12}{132} \\ P(B_1, B_2) &= \frac{3}{12} \times \frac{2}{11} = \frac{6}{132} \end{align*}

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: (R1,R2)(R_1, R_2), (W1,R2)(W_1, R_2), (B1,R2)(B_1, R_2).

P(shangayamwishoninyekundu)=P(R1,R2)+P(W1,R2)+P(B1,R2)P(shanga ya mwisho ni nyekundu) = P(R_1, R_2) + P(W_1, R_2) + P(B_1, R_2) =20132+20132+15132=55132= \frac{20}{132} + \frac{20}{132} + \frac{15}{132} = \frac{55}{132}

Kurahisisha sehemu:

\frac{55}{132} = \frac{55 \div 11}{132 \div 11} = \frac{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: (R1,R2)(R_1, R_2), (W1,W2)(W_1, W_2), (B1,B2)(B_1, B_2).

P(rangimoja)=P(R1,R2)+P(W1,W2)+P(B1,B2)P(rangi moja) = P(R_1, R_2) + P(W_1, W_2) + P(B_1, B_2) =20132+12132+6132=38132= \frac{20}{132} + \frac{12}{132} + \frac{6}{132} = \frac{38}{132}

Kurahisisha sehemu:

\frac{38}{132} = \frac{38 \div 2}{132 \div 2} = \frac{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)=95 (nyekundu) + 4 (nyepe) = 9.

P(hakunashangayabluu)=P(yakwanzasibluu)×P(yapilisibluu∣yakwanzasibluu)P(hakuna shanga ya bluu) = P(ya kwanza si bluu) \times P(ya pili si bluu | ya kwanza si bluu) =912×811=72132= \frac{9}{12} \times \frac{8}{11} = \frac{72}{132}

Kurahisisha sehemu:

72132=72÷12132÷12=611\frac{72}{132} = \frac{72 \div 12}{132 \div 12} = \frac{6}{11}

Sasa, P(angalau moja ni bluu) = 1−P(hakunashangayabluu)1 - P(hakuna shanga ya bluu)

= 1 - \frac{6}{11} = \frac{11}{11} - \frac{6}{11} = \frac{5{11}}
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