Probability of rumors

Question

A rumor is spread randomly among a group of $10$ people by successively having one person call someone, who calls someone, and so on. A person can pass the rumor on to anyone except the individual who just called.

• (a) What is the probability that if $A$ starts the rumor, $A$ receives the third call?
  
• (b) What is the probability that if $A$ does not start the rumor, $A$ receives the third call?

My attempt:

• (a) Total ways in which third call can be received is: $1$ (ways in which rumor can start)$\times 9$ (first call)$\times 8$ (second call)$\times 8$ (third call)

  Ways in which third call can be received by $A$ is: $1$ (ways in which rumor can start)$\times 9$ (first call)$\times 8$ (second call)$\times 1$ (third call received by $A$)

  So, probability $= (1\times9\times8\times1)/(1\times9\times8\times8)$

• (b) Total ways in which third call can be received is: $9$ (ways in which rumor can start)$\times 9$ (first call)$\times 8$ (second call)$\times 8$ (third call)

  Ways in which third call can be received by $A$ is: $9$ (ways in which rumor can start)$\times 9$ (first call)$\times 8$ (second call)$\times 1$ (third call received by $A$)
  

  So, probability $= (9\times9\times8\times1)/(9\times9\times8\times8)$

Please tell me if I am correct.

Thanks in advance!

Answer 1

(a) is correct.

In (b), the numerator should be $9\times 8\times 7$. Suppose the rumor starts at $X$ and goes to $Y$ then $Z$ then $A$. There are $9$ choices for $X$, as you said. Note $Y$ cannot equal $A$, otherwise $Z$ would be calling the person they just called. Therefore, there are only $8$ choices for $Y$. Finally, there are $7$ choices for $Z$, as $Z$ cannot equal $X$ oy $Y$ or $A$.