Solveeit Logo

Question

Question: A contest consists of predicting the results (win, draw or defeat) of 10 football matches. The proba...

A contest consists of predicting the results (win, draw or defeat) of 10 football matches. The probability that one entry contains at least 5 correct answers is
(A) 12585310\dfrac{12585}{{{3}^{10}}}
(B) 12385310\dfrac{12385}{{{3}^{10}}}
(C) 9335310\dfrac{9335}{{{3}^{10}}}
(D) 12096310\dfrac{12096}{{{3}^{10}}}

Explanation

Solution

Hint: Possible results of a match can be any of win, draw or defeat. So, for a match, only one possibility will be correct and the rest two will be wrong. The probability for correct prediction is 13\dfrac{1}{3} . The probability of the wrong prediction is 23\dfrac{2}{3} . For exactly n correct answers out of 10 matches, we have the formula, 10Cn(correctprediction)n(wrongprediction)10n^{10}{{C}_{n}}{{(correct\,prediction)}^{n}}{{(wrong\,prediction)}^{10-n}} . Use this formula and find the summation of the probability of exactly 5 correct predictions, exactly 6 correct predictions, exactly 7 correct predictions, exactly 8 correct predictions, exactly 9 correct predictions, and exactly 10 correct predictions.

Complete step-by-step answer:
The result of a match can be a win, draw, or defeat. The outcome can be only one of win, draw or defeat, rest two predictions will be wrong.
Probability for correct prediction = 13\dfrac{1}{3} .
Probability for wrong prediction = 23\dfrac{2}{3} .
We have to find the probability for at least 5 correct answers.
Probability for exactly 5 correct answers = 10C5(13)5(23)5^{10}{{C}_{5}}{{\left( \dfrac{1}{3} \right)}^{5}}{{\left( \dfrac{2}{3} \right)}^{5}} ………………………..(1)
Probability for exactly 6 correct answers = 10C6(13)6(23)4^{10}{{C}_{6}}{{\left( \dfrac{1}{3} \right)}^{6}}{{\left( \dfrac{2}{3} \right)}^{4}} …………………………(2)
Probability for exactly 7 correct answers = 10C7(13)7(23)3^{10}{{C}_{7}}{{\left( \dfrac{1}{3} \right)}^{7}}{{\left( \dfrac{2}{3} \right)}^{3}} …………………………(3)
Probability for exactly 8 correct answers = 10C8(13)8(23)2^{10}{{C}_{8}}{{\left( \dfrac{1}{3} \right)}^{8}}{{\left( \dfrac{2}{3} \right)}^{2}} …………………………(4)
Probability for exactly 9 correct answers = 10C9(13)9(23)1^{10}{{C}_{9}}{{\left( \dfrac{1}{3} \right)}^{9}}{{\left( \dfrac{2}{3} \right)}^{1}} …………………………(5)
Probability for exactly 10 correct answers = 10C10(13)10(23)0^{10}{{C}_{10}}{{\left( \dfrac{1}{3} \right)}^{10}}{{\left( \dfrac{2}{3} \right)}^{0}} …………………………(6)
Probability for at least 5 correct answers = Probability for exactly 5 correct answers + Probability for exactly 6 correct answers + Probability for exactly 7 correct answers + Probability for exactly 8 correct answers + Probability for exactly 9 correct answers + Probability for exactly 10 correct answers ………………………(7)
From equation (1), equation (2), equation (3), equation (4), equation (5), equation (6) and equation (7), we get
Probability for at least 5 correct answers = 10C5(13)5(23)5^{10}{{C}_{5}}{{\left( \dfrac{1}{3} \right)}^{5}}{{\left( \dfrac{2}{3} \right)}^{5}} + 10C6(13)6(23)4^{10}{{C}_{6}}{{\left( \dfrac{1}{3} \right)}^{6}}{{\left( \dfrac{2}{3} \right)}^{4}} + 10C7(13)7(23)3^{10}{{C}_{7}}{{\left( \dfrac{1}{3} \right)}^{7}}{{\left( \dfrac{2}{3} \right)}^{3}} + 10C8(13)8(23)2^{10}{{C}_{8}}{{\left( \dfrac{1}{3} \right)}^{8}}{{\left( \dfrac{2}{3} \right)}^{2}} + 10C9(13)9(23)1^{10}{{C}_{9}}{{\left( \dfrac{1}{3} \right)}^{9}}{{\left( \dfrac{2}{3} \right)}^{1}} + 10C10(13)10(23)0^{10}{{C}_{10}}{{\left( \dfrac{1}{3} \right)}^{10}}{{\left( \dfrac{2}{3} \right)}^{0}} …………………(8)
10C5=10.9.8.7.61.2.3.4.5=252^{10}{{C}_{5}}=\dfrac{10.9.8.7.6}{1.2.3.4.5}=252
10C6=10.9.8.7.6.51.2.3.4.5.6=210^{10}{{C}_{6}}=\dfrac{10.9.8.7.6.5}{1.2.3.4.5.6}=210
10C7=10.9.8.7.6.5.41.2.3.4.5.6.7=120^{10}{{C}_{7}}=\dfrac{10.9.8.7.6.5.4}{1.2.3.4.5.6.7}=120
10C8=10.9.8.7.6.5.4.31.2.3.4.5.6.7.8=45^{10}{{C}_{8}}=\dfrac{10.9.8.7.6.5.4.3}{1.2.3.4.5.6.7.8}=45
10C9=10.9.8.7.6.5.4.3.21.2.3.4.5.6.7.8.9=10^{10}{{C}_{9}}=\dfrac{10.9.8.7.6.5.4.3.2}{1.2.3.4.5.6.7.8.9}=10
10C10=10.9.8.7.6.5.4.3.2.11.2.3.4.5.6.7.8.9.10=1^{10}{{C}_{10}}=\dfrac{10.9.8.7.6.5.4.3.2.1}{1.2.3.4.5.6.7.8.9.10}=1
Putting 10C5=252^{10}{{C}_{5}}=252 , 10C6=210^{10}{{C}_{6}}=210 , 10C7=120^{10}{{C}_{7}}=120 , 10C8=45^{10}{{C}_{8}}=45 , 10C9=10^{10}{{C}_{9}}=10 , and 10C10=1^{10}{{C}_{10}}=1 in equation (8). On solving we get,
Probability for at least 5 correct answers = 32(252)310+16(210)310+8(120)310+4(45)310+2(10)310+1.1310\dfrac{32(252)}{{{3}^{10}}}+\dfrac{16(210)}{{{3}^{10}}}+\dfrac{8(120)}{{{3}^{10}}}+\dfrac{4(45)}{{{3}^{10}}}+\dfrac{2(10)}{{{3}^{10}}}+\dfrac{1.1}{{{3}^{10}}}

& =\dfrac{8064}{{{3}^{10}}}+\dfrac{3360}{{{3}^{10}}}+\dfrac{960}{{{3}^{10}}}+\dfrac{180}{{{3}^{10}}}+\dfrac{20}{{{3}^{10}}}+\dfrac{1}{{{3}^{10}}} \\\ & =\dfrac{8064+3360+960+180+20+1}{{{3}^{10}}} \\\ & =\dfrac{12585}{{{3}^{10}}} \\\ \end{aligned}$$ So, the probability for at least 5 correct answers is $$\dfrac{12585}{{{3}^{10}}}$$ . Hence, option (A) is correct. Note: In this question, one can find the probability for exactly 5 correct answers and then conclude the wrong answer. We have to find the probability for at least 5 correct answers. It means that we have to find the probability that the minimum 5 answers should be correct. There are no restrictions on the maximum number of correct answers. The maximum correct answers can be any of 6, 7, 8, 9, and 10. So, Probability for at least 5 correct answers = Probability for exactly 5 correct answers + Probability for exactly 6 correct answers + Probability for exactly 7 correct answers + Probability for exactly 8 correct answers + Probability for exactly 9 correct answers + Probability for exactly 10 correct answers.