Question

A football match may be either won, drawn or lost by the host country's team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.

Answer 1

According to the given question, firstly we will calculate the probability of the correct and incorrect result using the formula of probability = $\dfrac{\text{favourable outcomes}}{\text{Total number of outcomes}}$. Then calculate the probability of forecasting at least three correct results for four matches using the given and calculated values.

Formula used:
Here we use the formula of probability that is $\dfrac{\text{favourable outcomes}}{\text{Total number of outcomes}}$

Complete step by step solution:
It is given that there are three ways of forecasting the result of any one of the football matches.
As, there are one correct and two incorrect.
Now, we will calculate the probability of the correct result.
As here, the total number of outcomes is the sum of correct and incorrect that is $1 + 2 = 3$ .
So, favourable outcomes is equal to the correct result that is 1.
So, using the formula of probability = $\dfrac{{favourable\,{\rm{ }}outcomes}}{{Total\,{\rm{ }}number\,{\rm{ }}\,of{\rm{ }}\,outcomes}}$
After substituting the values we get,
The probability of correct result = $\dfrac{1}{3}$
And similarly, we are also calculating the probability of incorrect result using the above formula we get,
The probability of incorrect result = $\dfrac{2}{3}$
Now, we will calculate the probability of forecasting at least three correct for four matches.
Therefore, probability = $4 \times {\left( {\dfrac{1}{3}} \right)^3} \times \dfrac{2}{3} + {\left( {\dfrac{1}{3}} \right)^4}$
On simplifying we get,
$\Rightarrow 4 \times \dfrac{1}{{27}} \times \dfrac{2}{3} + \dfrac{1}{{81}}$
Further multiplying and by taking L.C.M we get,
$\Rightarrow \dfrac{8}{{81}} + \dfrac{1}{{81}}$
$\Rightarrow \dfrac{9}{{81}}$
On dividing with 9 from both numerator and denominator:
So, we get
$\Rightarrow \dfrac{1}{9}$

Hence, the probability of forecasting at least three correct for four matches = $\dfrac{1}{9}$

Note:
To solve these types of questions, we also have an alternative method that is by using binomial probability. You can use the relation $\sum\limits_{k = 0}^n {{C_{n,k}}{{(p)}^k}{{(1 - p)}^{n - k}} = 1}$ where $n = 4$ that is total number of matches. Then calculate the probability of the correct and incorrect result as shown above. Then substitute all the values in the given formula. Hence, calculate the probability.