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

Hint- In order to solve the problem, first we will separately find the probability of correct forecasting of result and incorrect forecasting of result for one particular match then we will use the same data to find the overall probability of correct prediction of at least three results.

Complete step-by-step answer:
Given that: There are three ways of forecasting the result of any one match, one correct and two incorrect.
As only one out of three predictions can be done.
So, Probability of correct result
$ = \dfrac{1}{3}$
And Probability of incorrect result
$ = \dfrac{2}{3}$
Now, we have to find the probability of forecasting at least three correct results for four matches.
So, probability of forecasting at least three correct results for four matches = Probability of 3 correct results and 1 incorrect result or probability of all 4 correct results.
Let us separately find both of them.
Probability of 3 correct result and 1 incorrect result:
For three correct results we have \[\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} = {\left( {\dfrac{1}{3}} \right)^3}\]
For one incorrect result we have \[\dfrac{2}{3}\]
Also this incorrect result can be any of the 4 matches.
So, Probability of 3 correct result and 1 incorrect result
\[
   = \left( 4 \right) \times \left( {\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3}} \right) \times \left( {\dfrac{2}{3}} \right) \\
   = 4 \times {\left( {\dfrac{1}{3}} \right)^3} \times \dfrac{2}{3} \\
 \]
Probability of 4 correct result:
For four correct results we have \[\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} = {\left( {\dfrac{1}{3}} \right)^4}\]
Now let us find the overall probability of forecasting at least three correct results for four matches.
= Probability of 3 correct results and 1 incorrect result or probability of all 4 correct results.
$ = \left( {4 \times {{\left( {\dfrac{1}{3}} \right)}^3} \times \dfrac{2}{3}} \right) + \left( {{{\left( {\dfrac{1}{3}} \right)}^4}} \right)$
Now let us simplify the term to find out the probability by taking terms common.
\[
   = {\left( {\dfrac{1}{3}} \right)^3} \times \left( {\left( {4 \times \dfrac{2}{3}} \right) + \left( {\dfrac{1}{3}} \right)} \right) \\
   = {\left( {\dfrac{1}{3}} \right)^3} \times \left( {\dfrac{8}{3} + \dfrac{1}{3}} \right) \\
   = {\left( {\dfrac{1}{3}} \right)^3} \times \left( {\dfrac{9}{3}} \right) \\
   = {\left( {\dfrac{1}{3}} \right)^3} \times 3 \\
   = \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} \times 3 \\
   = \dfrac{1}{3} \times \dfrac{1}{3} \\
   = \dfrac{1}{9} \\
 \]
Hence, the probability of forecasting at least three correct results for four matches is $\dfrac{1}{9}$ .

Note- The case shown above is an example of mutually exclusive events. Two events are mutually exclusive if they cannot occur at the same time. If two events are mutually exclusive, then the probability of either occurring is the sum of the probabilities of each occurring. The rule used above is a specific addition rule, where we find the sum of two probabilities.