Question

A contest consists of predicting the results namely (draw, win, defeat) of 20 cricket matches. The number of ways in which entry contains at least 5 correct answers is
A. $ {3^{20}}\sum\limits_{r = 0}^{20} {{}^{20}{C_r}{2^r}} $
B. $ {3^{20}} - \sum\limits_{r = 0}^4 {{}^{20}{C_r}{2^{20 - r}}} $
C. $ \sum\limits_{r = 1}^5 {{}^{20}{C_r}} $
D. $ \sum\limits_{r = 6}^{20} {{}^{10}{C_r}{3^r}} $

Answer 1

Hint: There are 3 possible outcomes which are draw, win and defeat and 20 trials (cricket matches). So the total possible outcomes can be $ {3^{20}} $ . To get the no. of ways the entry contains at least 5 correct answers, we have to subtract the (no. of ways the entry contains 0, 1, 2, 3, 4 correct answers) from the (total no. of ways).

Complete step-by-step answer:
We are given that a contest consists of predicting the results namely (draw, win, defeat) of 20 cricket matches.
We have to find the no. of ways in which entry contains at least 5 correct answers.
The no. of ways the entry contains at least 5 correct answers is equal to total possible outcomes minus no. of ways the entry contains 0, 1, 2, 3, 4 correct answers (less than 5 correct answers).
When the match is drawn it cannot be win and defeat, when it is win it cannot be draw and defeat and so on, this means when there is 1 success then there are 2 failures.
Therefore, the Total number of outcomes $ {3^{20}} $ can be divided as
$\Rightarrow {}^{20}{C_0}{2^{20}} + {}^{20}{C_1}{2^{19}} + {}^{20}{C_2}{2^{18}} + ...... + {}^{20}{C_{20}}{2^0} $
 Number of ways entry contains 0, 1, 2, 3, 4 correct answers is
$\Rightarrow {}^{20}{C_0}{1^0}{.2^{20}} + {}^{20}{C_1}{1^1}{.2^{19}} + {}^{20}{C_2}{1^2}{.2^{18}} + {}^{20}{C_3}{1^3}{.2^{17}} + {}^{20}{C_4}{1^4}{.2^{16}} = \sum\limits_{r = 0}^4 {{}^{20}{C_r}{1^r}{{.2}^{20 - r}} = } \sum\limits_{r = 0}^4 {{}^{20}{C_r}{2^{20 - r}}} $
The no. of ways the entry contains at least 5 correct answers is
$\Rightarrow {3^{20}} - \sum\limits_{r = 0}^4 {{}^{20}{C_r}{2^{20 - r}}} $
The correct option is Option B, $ {3^{20}} - \sum\limits_{r = 0}^4 {{}^{20}{C_r}{2^{20 - r}}} $ .

So, the correct answer is “Option B”.

Note: We know that the expansion of $ {\left( {x + y} \right)^n} $ using binomial expansion is $ {}^n{C_0}{x^0}{y^n} + {}^n{C_1}{x^1}{y^{n - 1}} + {}^n{C_2}{x^2}{y^{n - 2}} + .... + {}^n{C_n}{x^n}{y^0} $
When the value of x is 1, y is 2 and n is 20, then
  $
\Rightarrow {\left( {1 + 2} \right)^{20}} = {}^{20}{C_0}{1^0}{2^{20}} + {}^{20}{C_1}{1^1}{2^{19}} + {}^{20}{C_2}{1^2}{2^{18}} + .... + {}^{20}{C_{20}}{1^{20}}{2^0} \\
   \Rightarrow {3^{20}} = {}^{20}{C_0}{2^{20}} + {}^{20}{C_1}{2^{19}} + {}^{20}{C_2}{2^{18}} + ...... + {}^{20}{C_{20}}{2^0} \\
  $