If 10 cars are parked a row, then what  is the probability that there are exactly 5 cars between 2 particular cars

 

  • Option 1)

    \frac{8\times 5!}{10!}

  • Option 2)

    \frac{4\times 5!\times 3!\times 2!}{10!}

  • Option 3)

    \frac{8\times 8!}{10!}

  • Option 4)

    \frac{16\times 8!}{10!}

 

Answers (1)

 

Probability of occurrence of an event -

Let S be the sample space then the probability of occurrence of an event E is denoted by P(E) and it is defined as 

P\left ( E \right )=\frac{n\left ( E \right )}{n\left ( S \right )}

P\left ( E \right )\leq 1

P(E)=\lim_{n\rightarrow\infty}\left(\frac{r}{n} \right )

 

 

- wherein

Where n repeated experiment and E occurs r times.

 

 Since 2 cars are fix so there are eight places. and they can arranged by 8! Now these two cars also changed by eight places [4X2=8] so that chance 8X8!

So required probability 

 

\frac{8\times 8!}{10!}

 


Option 1)

\frac{8\times 5!}{10!}

This option is incorrect

Option 2)

\frac{4\times 5!\times 3!\times 2!}{10!}

This option is incorrect

Option 3)

\frac{8\times 8!}{10!}

This option is correct

Option 4)

\frac{16\times 8!}{10!}

This option is incorrect

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