The number of finishers at a particularly difficult race is a random variable with probability mass function: $$f ( x ) = \left\{ \begin{array} { l l }
\frac { 36 } { 35 ( x ) ( x + 1 ) } & x = 1 , \ldots , 35 \
0 & \text { else. }
\end{array} \right.$$ Find the probability that at least 3 people do not finish.

Question

The number of finishers at a particularly difficult race is a random variable with probability mass function: $$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 36 } { 35 ( x ) ( x + 1 ) } & x = 1 , \ldots , 35 \
0 & \text { else. }
\end{array} \right.$$ Find the probability that at least 3 people do not finish.

Quizplus · Accepted Answer

The Answer of The number of finishers at a particularly difficult race is a random variable with probability mass function: $$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 36 } { 35 ( x ) ( x + 1 ) } & x = 1 , \ldots , 35 \
0 & \text { else. }
\end{array} \right.$$ Find the probability that at least 3 people do not finish.

The Number of Finishers at a Particularly Difficult Race Is f(x)={3635(x)(x+1)x=1,…,350 else. f ( x ) = \left\{ \begin{array} { l l } \frac { 36 } { 35 ( x ) ( x + 1 ) } & x = 1 , \ldots , 35 \\0 & \text { else. }\end{array} \right.f(x)={35(x)(x+1)36​0​x=1,…,35 else. ​

The Number of Finishers at a Particularly Difficult Race Is $f ( x ) = \left\{ \begin{array} { l l } \frac { 36 } { 35 ( x ) ( x + 1 ) } & x = 1 , \ldots , 35 \\0 & \text { else. }\end{array} \right.$