Question

Three numbers are chosen from 1 to 15. Find the probability that they are consecutive.

Answer 1

Hint: Given that, three numbers are chosen from 1 to 15. Therefore, total number of ways of selecting 3 numbers from 15 numbers is\[{}^{15}C{_3}\]
 Now find the number of outcomes where 3 numbers are consecutive.
 Therefore, required probability will be
$ = \dfrac{{{\text{number of outcomes where three numbers are consecutive}}}}{{{\text{total number of ways of choosing three numbers from 1 to 15}}}}$

Complete step-by-step answer:
Given that, three numbers are chosen from 1 to 15.
The total number of ways of selecting 3 numbers from 15 numbers is
 \[ = \;{}^{15}C{_3}\]
 Using, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
 \[ = \dfrac{{15!}}{{3\,!\,\, \times 12!}}\]
 On simplification we get,
\[ = \dfrac{{15 \times 14 \times 13}}{6}\]
 On solving we get,
 \[ = 455\]
 Now we have to find the probability that the three numbers are consecutive.
Therefore, The favourable outcomes are $\left\{ {\left( {1,2,3} \right),\,\left( {2,3,4} \right),\,\,...\,,\,\left( {13,14,15} \right)} \right\}$
 Hence there are 13 possible outcomes.
 Now, \[{\text{probability of an event happening = p(E) = }}\dfrac{{{\text{Number of ways it can happen}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{n(E)}}{{n(S)}}\]
 Therefore, required probability is
$ = \dfrac{{{\text{number of outcomes where three numbers are consecutive}}}}{{{\text{total number of ways of choosing three numbers from 1 to 15}}}}$
 On substituting the values, we get
 $ = \dfrac{{13}}{{455}}$
 On simplification we get,
 $ = \dfrac{1}{{35}}$
Hence, probability is $\dfrac{1}{{35}}$.

Note: Points to remember, \[{\text{probability of an event happening = p(E) = }}\dfrac{{{\text{Number of ways it can happen}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{n(E)}}{{n(S)}}\]
Again, total number of ways of selecting r numbers from n numbers is \[{}^nC{_r}\]; where ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$