Question

Mr. A forgot to write down a very important phone number. All he remembers is that it started with $713$ and the next set of $4$ digits involved are $1,7$ and $9$ with one of these numbers appearing twice. He guessed 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 $1/35$

Let total number of possible events be $P\left(n\right)$

$P\left(n\right)={}^4{P}_2\times {}^3{C}_1$

Where ${}^4{P}_2$ is total events for arranging three numbers on four blank spaces when one number repeats and ${}^3{C}_1$ is total ways of choosing the repeating number out of $1,7$ and $9$.

Product of both is taken to find total $P\left(n\right)$ because both execute simultaneously.

Successful event will be $P\left(m\right)=1$

Now, the odds in favor of dialing the correct number

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

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

$=1:35$