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) 31012585
(B) 31012385
(C) 3109335
(D) 31012096
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 31 . The probability of the wrong prediction is 32 . For exactly n correct answers out of 10 matches, we have the formula, 10Cn(correctprediction)n(wrongprediction)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 = 31 .
Probability for wrong prediction = 32 .
We have to find the probability for at least 5 correct answers.
Probability for exactly 5 correct answers = 10C5(31)5(32)5 ………………………..(1)
Probability for exactly 6 correct answers = 10C6(31)6(32)4 …………………………(2)
Probability for exactly 7 correct answers = 10C7(31)7(32)3 …………………………(3)
Probability for exactly 8 correct answers = 10C8(31)8(32)2 …………………………(4)
Probability for exactly 9 correct answers = 10C9(31)9(32)1 …………………………(5)
Probability for exactly 10 correct answers = 10C10(31)10(32)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(31)5(32)5 + 10C6(31)6(32)4 + 10C7(31)7(32)3 + 10C8(31)8(32)2 + 10C9(31)9(32)1 + 10C10(31)10(32)0 …………………(8)
10C5=1.2.3.4.510.9.8.7.6=252
10C6=1.2.3.4.5.610.9.8.7.6.5=210
10C7=1.2.3.4.5.6.710.9.8.7.6.5.4=120
10C8=1.2.3.4.5.6.7.810.9.8.7.6.5.4.3=45
10C9=1.2.3.4.5.6.7.8.910.9.8.7.6.5.4.3.2=10
10C10=1.2.3.4.5.6.7.8.9.1010.9.8.7.6.5.4.3.2.1=1
Putting 10C5=252 , 10C6=210 , 10C7=120 , 10C8=45 , 10C9=10 , and 10C10=1 in equation (8). On solving we get,
Probability for at least 5 correct answers = 31032(252)+31016(210)+3108(120)+3104(45)+3102(10)+3101.1