In an examination of 9 papers,a candidate is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $n = 9$ be the total number of papers. Let $p$ be the number of papers passed and $f$ be the number of papers failed,where $p + f = 9$.The can

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(B) Let $n = 9$ be the total number of papers. Let $p$ be the number of papers passed and $f$ be the number of papers failed,where $p + f = 9$.The candidate is successful if $p > f$. Since $p = 9 - f$,the condition $p > f$ becomes $9 - f > f$,which implies $2f Thus,the candidate is successful if they fail in $0, 1, 2, 3,$ or $4$ papers.The candidate is unsuccessful if they fail in $5, 6, 7, 8,$ or $9$ papers.The number of ways to be unsuccessful is the sum of combinations: $\binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,this is equal to $\binom{9}{4} + \binom{9}{3} + \binom{9}{2} + \binom{9}{1} + \binom{9}{0}$.We know that the sum of all binomial coefficients $\sum_{k=0}^{9} \binom{9}{k} = 2^9 = 512$.Since $\sum_{k=0}^{9} \binom{9}{k} = (\binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4}) + (\binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9}) = 512$,and both sums are equal,each sum is $\frac{512}{2} = 256$.Therefore,the number of ways to be unsuccessful is $256$.

In an examination of $9$ papers,a candidate is successful only if they pass in more papers than they fail. In how many ways can the candidate be unsuccessful?

Similar Questions

In how many ways can a committee of $6$ members be formed out of $10$ members,such that it always includes a specified member?

Two players $A$ and $B$ play a series of games of badminton. The player, who wins $5$ games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player $A$ wins the series is . . . . . . .

In how many ways can $6$ '$+$' signs and $4$ '$$' signs be arranged in a row such that no two '$$' signs occur together?

The number of values of $r$ satisfying $^{69}C_{3r-1} - ^{69}C_{r^2} = ^{69}C_{r^2-1} - ^{69}C_{3r}$ is:

There are $10$ bulbs in a room. Each of them can be switched on independently. In how many ways can the room be illuminated?

Vedclass Test Series

Exam Paper Generator

Online Exam Module