Question

Mr. A forgot to write down a very important phone number. All he remembers is that it started with 713 and that the next set of 4 digit involved are 1,7 and 9 with one of these numbers appearing twice.He guesses a phone number and dials randomly. The odds in favour of dialing the correct telephone number, is

• A. $1:35$
• B. $1:71$
• C. $1:23$
• D. $1:36$

Answer 1

The correct option is A $1:35$

$n=$ sample space

Since, one of these numbers in the last $4$ digits is appearing twice

$P\left(n\right)=\frac{4!}{2!}{\times }^3{C}_1$

$=\frac{4\times 3\times 2!}{2!}\times \frac{3!}{2!\times 1!}$

$=36$

$P\left(m\right)=1$ success

Odds in favour = success / no. of unsuccessful

$=\frac{m}{n-m}$

$=\frac{P\left(m\right)}{P\left(n\right)-P\left(m\right)}$

$=\frac{1}{36-1}=\frac{1}{35}$