question_answer

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 i.e. one correct and two incorrect. Find the probability of forecasting atleast three correct results for four matches.

Answer:

Given, total number of ways of forecasting the result = 3 Number of ways of forecasting the correct result                         = 1 and number of ways of forecasting the incorrect result                         = 2 \[\therefore \] Probability of forecasting the correct result                            \[=\frac{1}{3}=P(S)(say)\] and probability of forecasting the incorrect result                         \[=1-\frac{1}{3}=\frac{2}{3}=P(F)(say)\] In four matches, probability of forecasting atleast three correct results      \[=P[(SSSF)\,\text{Or}\,(SFSS)\,\text{Or}\,(SSFS)\,\text{Or}\,(FSSS)\,\text{Or}\,(SSSS)]\]\[=P(S)P(S)P(S)P(F)+P(S)P(F)P(S)P(S)\]             \[+P(S)P(S)P(F)P(S)+P(F)P(S)P(S)P(S)\]                                          \[+P(S)P(S)P(S)P(S)\] \[=\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}\times \frac{2}{3}+\frac{1}{3}\times \frac{2}{3}\times \frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{3}\times \frac{2}{3}\times \frac{1}{3}\]                            \[+\frac{2}{3}\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}\] \[=\frac{2}{81}+\frac{2}{81}+\frac{2}{81}+\frac{2}{81}+\frac{1}{81}=\frac{2+2+2+2+1}{81}\] \[=\frac{9}{81}=\frac{1}{9}\]