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Make P the subject of the formula XY^r = Q^r

Mathematics

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Answer

P(R)=1960P(R) = \frac{19}{60}

Step-by-step solution

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Huu hapa ni utatuzi wa swali la 12.

Swali la 12: Mfuko mmoja una mipira 10 ambapo 3 ni mekundu, 5 ni meupe na 2 ni ya kijani. Mfuko mwingine una mipira 12 ambapo 4 ni mekundu, 3 ni meupe na 5 ni ya kijani. Mfuko unachaguliwa kwa nasibu na mpira unachaguliwa kwa nasibu. Tafuta uwezekano kwamba mpira uliochaguliwa ni mwekundu.

  1. Tambua idadi ya mipira na uwezekano wa kuchagua mpira mwekundu kutoka kila mfuko.

    • Mfuko wa Kwanza (B1B_1):

      • Jumla ya mipira = 10
      • Mipira mekundu = 3
      • Uwezekano wa kuchagua mpira mwekundu kutoka Mfuko wa Kwanza, P(R∣B1)=310P(R|B_1) = \frac{3}{10}.

    • Mfuko wa Pili (B2B_2):

      • Jumla ya mipira = 12
      • Mipira mekundu = 4
      • Uwezekano wa kuchagua mpira mwekundu kutoka Mfuko wa Pili, P(R∣B2)=412=13P(R|B_2) = \frac{4}{12} = \frac{1}{3}.

  2. Tambua uwezekano wa kuchagua kila mfuko.

    Kwa kuwa mfuko unachaguliwa kwa nasibu na kuna mifuko miwili:

    • Uwezekano wa kuchagua Mfuko wa Kwanza, P(B1)=12P(B_1) = \frac{1}{2}.
    • Uwezekano wa kuchagua Mfuko wa Pili, P(B2)=12P(B_2) = \frac{1}{2}.

  3. Tumia kanuni ya jumla ya uwezekano (Total Probability Rule) kupata uwezekano wa kuchagua mpira mwekundu.

    P(R)=P(R∣B1)P(B1)+P(R∣B2)P(B2)P(R) = P(R|B_1)P(B_1) + P(R|B_2)P(B_2)

    Badilisha thamani ulizopata:

    P(R)=(310)(12)+(13)(12)P(R) = \left(\frac{3}{10}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{3}\right)\left(\frac{1}{2}\right)

  4. Hesabu matokeo.

    P(R)=320+16P(R) = \frac{3}{20} + \frac{1}{6}

    Ili kujumlisha sehemu hizi, tafuta kigawo cha kawaida, ambacho ni 60.

    P(R)=3×320×3+1×106×10P(R) = \frac{3 \times 3}{20 \times 3} + \frac{1 \times 10}{6 \times 10} P(R)=960+1060P(R) = \frac{9}{60} + \frac{10}{60} P(R)=9+1060P(R) = \frac{9 + 10}{60} P(R)=1960P(R) = \boxed{\frac{19}{60}}
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