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Two independent relay teams want to qualify for the next Olympic Games. The probability that the two teams run under the qualifying time is 4/9 and 3/7 respectively. Calculate the probability that one of the relay teams will run under the qualifying time in their next race.

Mathematics

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Answer

3163\frac{31}{63}

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3.1.1: How many years ago did Sandile buy the car?

  1. Ngola foromo ea ho theoha ha boleng.

    Foromo ea ho theoha ha boleng ka mokhoa oa ho fokotsa chelete ke:

    A=P(1−i)nA = P(1 - i)^n

    moo AA e leng boleng ba hona joale, PP e leng theko ea pele, ii e leng sekhahla sa ho theoha ha boleng, 'me nn e le palo ea lilemo.

  2. Kenya boleng bo fanoeng ka har'a foromo.

    Re na le A=R79 866.96A = \text{R}79\,866.96, P=R180 000P = \text{R}180\,000, le i=15%=0.15i = 15\% = 0.15.

    79 866.96=180 000(1−0.15)n79\,866.96 = 180\,000(1 - 0.15)^n 79 866.96=180 000(0.85)n79\,866.96 = 180\,000(0.85)^n
  3. Rarolla bakeng sa nn.

    Arola mahlakore ka bobeli ka 180 000180\,000:

    79 866.96180 000=(0.85)n\frac{79\,866.96}{180\,000} = (0.85)^n 0.443705333...=(0.85)n0.443705333... = (0.85)^n

    Nka logarithm ea mahlakore ka bobeli ho rarolla bakeng sa nn:

    ln⁡(0.443705333...)=nln⁡(0.85)\ln(0.443705333...) = n \ln(0.85) n=ln⁡(0.443705333...)ln⁡(0.85)n = \frac{\ln(0.443705333...)}{\ln(0.85)} n≈5n \approx 5

    Sandile o rekile koloi lilemo tse 5 tse fetileng.

    3.1.2: Has Anil accumulated enough money in his savings account to buy Sandile's car now?

  4. Fumana nako ea letsete la Anil.

    Ho tsoa ho 3.1.1, Sandile o rekile koloi lilemo tse 5 tse fetileng. Anil o ile a beha chelete "ka nako e tšoanang le eo Sandile a rekileng koloi ka eona". Kahoo, nako ea letsete la Anil ke n=5n = 5 lilemo.

  5. Ngola foromo ea phaello e kopaneng.

    Foromo ea phaello e kopaneng ke:

    A=P(1+im)n×mA = P\left(1 + \frac{i}{m}\right)^{n \times m}

    moo AA e leng boleng ba nako e tlang, PP e leng chelete ea mantlha, ii e leng sekhahla sa phaello sa selemo le selemo, mm e leng palo ea linako tseo phaello e kopanngoang ka tsona ka selemo, 'me nn e le palo ea lilemo.

  6. Kenya boleng le ho bala chelete e bokelletsoeng ea Anil.

    Re na le P=R49 000P = \text{R}49\,000, i=10%=0.10i = 10\% = 0.10, m=4m = 4 (e kopanngoa kotara), le n=5n = 5 lilemo.

    A=49 000(1+0.104)5×4A = 49\,000\left(1 + \frac{0.10}{4}\right)^{5 \times 4} A=49 000(1+0.025)20A = 49\,000(1 + 0.025)^{20} A=49 000(1.025)20A = 49\,000(1.025)^{20} A≈49 000(1.63861644)A \approx 49\,000(1.63861644) A≈80 292.21A \approx 80\,292.21

    Anil o bokelletse R80 292.21.

  7. Bapisa chelete e bokelletsoeng ea Anil le boleng ba hona joale ba koloi ea Sandile.

    Boleng ba hona joale ba koloi ea Sandile ke R79 866.96.
    Chelete e bokelletsoeng ea Anil ke R80 292.21.
    Kaha R80 292.21 > R79 866.96, Anil o bokelletse chelete e lekaneng.
    Karabo: E, Anil o bokelletse chelete e lekaneng.

    3.2.1: How much did Jane owe immediately after making her 6th repayment?

  8. Bala sekhahla sa phaello sa khoeli le khoeli.

    Sekhahla sa phaello sa selemo le selemo ke iannual=10.25%=0.1025i_{annual} = 10.25\% = 0.1025.
    Phaello e kopanngoa khoeli le khoeli, kahoo sekhahla sa phaello sa khoeli le khoeli ke:

    i=0.102512i = \frac{0.1025}{12}
  9. Ngola foromo ea chelete e setseng ea kalimo.

    Chelete e setseng (PkP_k) ka mor'a litefo tse kk e fanoa ke:

    Pk=P(1+i)k−x((1+i)k−1)iP_k = P(1+i)^k - \frac{x((1+i)^k - 1)}{i}

    moo PP e leng chelete ea pele ea kalimo, xx e leng tefo ea khoeli le khoeli, ii e leng sekhahla sa phaello sa khoeli le khoeli, 'me kk e le palo ea litefo tse entsoeng.

  10. Kenya boleng le ho bala chelete e setseng ka mor'a litefo tse 6.

    P=R800 000P = \text{R}800\,000
    x=R7 853.15x = \text{R}7\,853.15
    i=0.102512≈0.0085416667i = \frac{0.1025}{12} \approx 0.0085416667
    k=6k = 6

    P6=800 000(1+0.102512)6−7 853.15((1+0.102512)6−1)0.102512P_6 = 800\,000\left(1 + \frac{0.1025}{12}\right)^6 - \frac{7\,853.15\left(\left(1 + \frac{0.1025}{12}\right)^6 - 1\right)}{\frac{0.1025}{12}} P6=800 000(1.0085416667)6−7 853.15((1.0085416667)6−1)0.0085416667P_6 = 800\,000(1.0085416667)^6 - \frac{7\,853.15((1.0085416667)^6 - 1)}{0.0085416667} P6=800 000(1.0523038)−7 853.15(0.0523038)0.0085416667P_6 = 800\,000(1.0523038) - \frac{7\,853.15(0.0523038)}{0.0085416667} P6=841 843.04−410.90690.0085416667P_6 = 841\,843.04 - \frac{410.9069}{0.0085416667} P6=841 843.04−48 106.90P_6 = 841\,843.04 - 48\,106.90 P6=793 736.14P_6 = 793\,736.14

    Jane o ne a kolota R793 736.14 hang ka mor'a ho etsa tefo ea hae ea bo6.

    3.2.2: Calculate Jane's increased monthly payment in order to settle the loan in the original 20 years.

  11. Bala chelete e setseng ka mor'a khoeli ea bo9 (ka mor'a litefo tse 3 tse fosahetseng).

    Chelete e setseng ka mor'a tefo ea bo6 e ne e le P6=R793 736.14P_6 = \text{R}793\,736.14.
    Jane o ile a fosa litefo tsa likhoeli tsa 7, 8, le 9. Nakong ena, phaello e ile ea bokellana holim'a chelete e setseng.
    Chelete e setseng ka mor'a likhoeli tse 9 (P9P_9) e tla ba P6P_6 e kopanngoa bakeng sa likhoeli tse 3:

    P9=P6(1+0.102512)3P_9 = P_6\left(1 + \frac{0.1025}{12}\right)^3 P9=793 736.14(1.0085416667)3P_9 = 793\,736.14(1.0085416667)^3 P9=793 736.14(1.025816)P_9 = 793\,736.14(1.025816) P9≈814 270.81P_9 \approx 814\,270.81

    Chelete e setseng ka mor'a likhoeli tse 9 ke R814 270.81.

  12. Fumana palo e setseng ea litefo.

    Nako ea pele ea kalimo e ne e le lilemo tse 20, e leng 20×12=24020 \times 12 = 240 likhoeli.
    Jane o se a fetile likhoeli tse 9 (litefo tse 6 li entsoe, tse 3 li fosetsoe).
    Palo e setseng ea litefo ke nrem=240−9=231n_{rem} = 240 - 9 = 231 likhoeli.

  13. Bala tefo e ncha ea khoeli le khoeli.

    Re hloka ho fumana tefo e ncha ea khoeli le khoeli (xnewx_{new}) e tla lefa chelete e setseng P9P_9 ka holim'a likhoeli tse setseng tsa nremn_{rem} ka sekhahla sa phaello sa khoeli le khoeli ii.
    Foromo ea boleng ba hona joale ba annuity ke Pn=x[1−(1+i)−n]iP_n = \frac{x[1-(1+i)^{-n}]}{i}.
    Ho rarolla bakeng sa xx:

    xnew=P9⋅i1−(1+i)−nremx_{new} = \frac{P_9 \cdot i}{1 - (1+i)^{-n_{rem}}} xnew=814 270.81×0.1025121−(1+0.102512)−231x_{new} = \frac{814\,270.81 \times \frac{0.1025}{12}}{1 - \left(1 + \frac{0.1025}{12}\right)^{-231}} xnew=814 270.81×0.00854166671−(1.0085416667)−231x_{new} = \frac{814\,270.81 \times 0.0085416667}{1 - (1.0085416667)^{-231}} xnew=6956.8871−0.144889x_{new} = \frac{6956.887}{1 - 0.144889} xnew=6956.8870.855111x_{new} = \frac{6956.887}{0.855111} xnew≈8135.69x_{new} \approx 8135.69

    Tefo e eketsehileng ea khoeli le khoeli ea Jane e tla ba R8 135.69.

    QUESTION 4: Calculate the probability that one of the relay teams will run under the qualifying time in their next race.

  14. Hlalosa menyetla bakeng sa sehlopha ka seng.

    E re P(T1)P(T_1) e be monyetla oa hore Sehlopha sa 1 se tla matha ka tlas'a nako e tšoanelehang.

    P(T1)=49P(T_1) = \frac{4}{9}

    E re P(T2)P(T_2) e be monyetla oa hore Sehlopha sa 2 se tla matha ka tlas'a nako e tšoanelehang.

    P(T2)=37P(T_2) = \frac{3}{7}
  15. Bala menyetla ea hore sehlopha ka seng se ke ke sa matha ka tlas'a nako e tšoanelehang.

    P(T1′)=1−P(T1)=1−49=59P(T_1') = 1 - P(T_1) = 1 - \frac{4}{9} = \frac{5}{9} P(T2′)=1−P(T2)=1−37=47P(T_2') = 1 - P(T_2) = 1 - \frac{3}{7} = \frac{4}{7}
  16. Bala monyetla oa hore hantle sehlopha se le seng sa lihlopha tsa relay se tla matha ka tlas'a nako e tšoanelehang.

    Sena se bolela hore Sehlopha sa 1 se atleha HAHOLO Sehlopha sa 2 se hloleha, KAPA Sehlopha sa 1 se hloleha HAHOLO Sehlopha sa 2 se atleha. Kaha liketsahalo li ikemetse, re ka atisa menyetla ea tsona.

    P(hantle sehlopha se le seng)=P(T1 le T2′)+P(T1′ le T2)P(\text{hantle sehlopha se le seng}) = P(T_1 \text{ le } T_2') + P(T_1' \text{ le } T_2) P(hantle sehlopha se le seng)=(P(T1)×P(T2′))+(P(T1′)×P(T2))P(\text{hantle sehlopha se le seng}) = (P(T_1) \times P(T_2')) + (P(T_1') \times P(T_2)) P(hantle sehlopha se le seng)=(49×47)+(59×37)P(\text{hantle sehlopha se le seng}) = \left(\frac{4}{9} \times \frac{4}{7}\right) + \left(\frac{5}{9} \times \frac{3}{7}\right) P(hantle sehlopha se le seng)=1663+1563P(\text{hantle sehlopha se le seng}) = \frac{16}{63} + \frac{15}{63} P(hantle sehlopha se le seng)=16+1563P(\text{hantle sehlopha se le seng}) = \frac{16 + 15}{63} P(hantle sehlopha se le seng)=3163P(\text{hantle sehlopha se le seng}) = \frac{31}{63}

    Monyetla oa hore sehlopha se le seng sa lihlopha tsa relay se tla matha ka tlas'a nako e tšoanelehang ke 3163\boxed{\frac{31}{63}}.

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