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The probability that wins two games. It either draws in the two matches or losses in both. It draws in one of the two matches. It wins in only one match.

Mathematics

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Answers

3 parts
a

536\frac{5}{36}

b

49\frac{4}{9}

c

12\frac{1}{2}

Step-by-step solution

7 steps

Huu hapa ni ufumbuzi wa swali lako. Tafadhali kumbuka kuwa uwezekano wa kushinda, sare, au kupoteza mchezo mmoja haukupewa kwenye swali. Kwa hivyo, nimechukua dhana zifuatazo kwa uwezekano wa mchezo mmoja:

  • Uwezekano wa kushinda mchezo mmoja, P(W)=12P(W) = \frac{1}{2}
  • Uwezekano wa sare katika mchezo mmoja, P(D)=13P(D) = \frac{1}{3}
  • Uwezekano wa kupoteza mchezo mmoja, P(L)=16P(L) = \frac{1}{6} Jumla ya uwezekano huu ni P(W)+P(D)+P(L)=12+13+16=3+2+16=66=1P(W) + P(D) + P(L) = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1. Ikiwa uwezekano halisi ni tofauti, tafadhali nijulishe.

Part a· 3 steps

Uwezekano kwamba timu itatoka sare katika mechi zote mbili au kupoteza katika zote mbili.

  1. Tafuta uwezekano wa sare katika mechi zote mbili.

    Kwa kuwa mechi ni huru, uwezekano wa sare katika mechi zote mbili ni P(D na D)=P(D)×P(D)P(D \text{ na } D) = P(D) \times P(D).

    P(D na D)=13×13=19P(D \text{ na } D) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

  2. Tafuta uwezekano wa kupoteza katika mechi zote mbili.

    Uwezekano wa kupoteza katika mechi zote mbili ni P(L na L)=P(L)×P(L)P(L \text{ na } L) = P(L) \times P(L).

    P(L na L)=16×16=136P(L \text{ na } L) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}

  3. Jumlisha uwezekano kwa matukio yanayotengana.

    Uwezekano wa sare katika mechi zote mbili AU kupoteza katika zote mbili ni jumla ya uwezekano wa matukio haya mawili.

    P(sare zote au kupoteza zote)=P(D na D)+P(L na L)P(\text{sare zote au kupoteza zote}) = P(D \text{ na } D) + P(L \text{ na } L) P(sare zote au kupoteza zote)=19+136=436+136=536P(\text{sare zote au kupoteza zote}) = \frac{1}{9} + \frac{1}{36} = \frac{4}{36} + \frac{1}{36} = \frac{5}{36}

    Uwezekano ni:

    536\boxed{\frac{5}{36}}

Part b

49\frac{4}{9}

Uwezekano kwamba timu itatoka sare katika moja ya mechi mbili.

  1. Tafuta uwezekano wa kutotoka sare katika mchezo mmoja.

    Uwezekano wa kutotoka sare ni P(sio D)=1−P(D)=1−13=23P(\text{sio D}) = 1 - P(D) = 1 - \frac{1}{3} = \frac{2}{3}.
    Hii inajumuisha kushinda au kupoteza.

  2. Tafuta uwezekano wa sare katika mechi moja tu.

    Kuna njia mbili za kutokea: (Sare, Sio Sare) au (Sio Sare, Sare).

    P(sare katika moja)=P(D)×P(sio D)+P(sio D)×P(D)P(\text{sare katika moja}) = P(D) \times P(\text{sio D}) + P(\text{sio D}) \times P(D) P(sare katika moja)=(13×23)+(23×13)P(\text{sare katika moja}) = \left(\frac{1}{3} \times \frac{2}{3}\right) + \left(\frac{2}{3} \times \frac{1}{3}\right) P(sare katika moja)=29+29=49P(\text{sare katika moja}) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}

    Uwezekano ni:

    49\boxed{\frac{4}{9}}

Part c

12\frac{1}{2}

Uwezekano kwamba timu itashinda katika mechi moja tu.

  1. Tafuta uwezekano wa kutoshinda katika mchezo mmoja.

    Uwezekano wa kutoshinda ni P(sio W)=1−P(W)=1−12=12P(\text{sio W}) = 1 - P(W) = 1 - \frac{1}{2} = \frac{1}{2}.
    Hii inajumuisha sare au kupoteza.

  2. Tafuta uwezekano wa kushinda katika mechi moja tu.

    Kuna njia mbili za kutokea: (Shinda, Sio Shinda) au (Sio Shinda, Shinda).

    P(shinda katika moja)=P(W)×P(sio W)+P(sio W)×P(W)P(\text{shinda katika moja}) = P(W) \times P(\text{sio W}) + P(\text{sio W}) \times P(W) P(shinda katika moja)=(12×12)+(12×12)P(\text{shinda katika moja}) = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2}\right) P(shinda katika moja)=14+14=24=12P(\text{shinda katika moja}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}

    Uwezekano ni:

    12\boxed{\frac{1}{2}}
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