Question

A contest consists of predicting the results namely (Draw,wins,defeat) of 20 cricket matches. The number of way in which entry contains at least 5 correct answer is

• A. $3^{20}{\sum }_{r=0}^{20}{}^{20}{C}_r{2}^r$
• B. $3^{20}-{\sum }_{r=0}^4{}^{20}{C}_r{2}^{20-r}$
• C. ${\sum }_{r=1}^5{}^{20}{C}_r$
• D. ${\sum }_{r=6}^{20}{}^{10}{C}_r{3}^r$

Answer 1

The correct option is B $3^{20}-{\sum }_{r=0}^4{}^{20}{C}_r{2}^{20-r}$

Let's denote win,loss and draw by W,L and D respectively.
Since 3 possible answers(W or L or D) can be attributed to all 20 matches hence the
Total possible answers=$3\times 3\times .........20times=3^{20}$
Now we have to subtract the cases in which there are 0 or 1 or 2 or 3 or 4 answers correct.
Number of ways of this can be done =$\left(\begin{array}{c}20\\ 0\end{array}\right)\times 2^{20}+\left(\begin{array}{c}20\\ 1\end{array}\right)\times 2^{19}+\left(\begin{array}{c}20\\ 2\end{array}\right)\times 2^{18}+\left(\begin{array}{c}20\\ 3\end{array}\right)\times 2^{17}+\left(\begin{array}{c}20\\ 4\end{array}\right)\times 2^{16}={\sum }_{r=0}^4\left(\begin{array}{c}20\\ r\end{array}\right)\times 2^{20-r}$
Hence the final answer =$3^{20}-{\sum }_{r=0}^4\left(\begin{array}{c}20\\ r\end{array}\right)\times 2^{20-r}$