A football match may be either won,draw or lost by the host country's team. So there are 3 ways of forecasting the result of any one match. What is the probability of forecasting at least 3 out of 4 matches?

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A football match may be either won,draw or lost by the host country's team. So there are 3 ways of forecasting the result of any one match. What is the probability of forecasting at least 3 out of 4 matches?

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Answer 1 · 2018-05-30T07:47:55

$\frac{1}{9} = 11 . ¯ 1 %$ $1/9=11.bar1%$

Explanation:

We can solve this using binomial probability. When starting a question like this, I like to start with this relation:

$n \sum k = 0 C_{n, k} {(p)}^{k} {(1 - p)}^{n - k} = 1$ $sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1$

We have 4 matches: $n = 4$ $n=4$ .

The probability of forecasting a match correctly is $\frac{1}{3}$ $1/3$ (there are 3 possible ways of forecasting and only 1 way of being right).

We're looking at getting at least 3 correct, so that's $3 \leq k \leq 4$ $3<=k<=4$ :

$C_{4, 3} {(\frac{1}{3})}^{3} {(\frac{2}{3})}^{1} + C_{4, 4} {(\frac{1}{3})}^{4} {(\frac{2}{3})}^{0}$ $C_(4,3)(1/3)^3(2/3)^(1)+C_(4,4)(1/3)^4(2/3)^(0)$

$= 4 (\frac{2}{81}) + 1 (\frac{1}{81}) = \frac{9}{81} = \frac{1}{9}$ $=4(2/81)+1(1/81)=9/81=1/9$

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A football match may be either won,draw or lost by the host country's team. So there are 3 ways of forecasting the result of any one match. What is the probability of forecasting at least 3 out of 4 matches?

1 Answer

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