Calculating probability for three consecutive successes Original Question:
On a TV news channel, the evening news starts at same time every day. The probability that Mr Li gets home from work in time to watch the news is $0.3$
In a particular week of five working days, what is the probability that Mr Li gets home in time to watch the news on three consecutive days?
Attempt:
Possible arrangements with three consecutive successes in 5 trials:
SSS..
FSSS.
FFSSS
SFSSS

I calculated the respective probabilities for these arrangements as follows:
$0.3^3 + 0.3^3 0.7 + 0.3^3 0.7^2+0.3^4 0.7 = 0.0648$
Actual answer (at the back of the book) = $0.05913$
My question is what I am doing wrong here (or is the book wrong?)
Also, I would really appreciate if someone could tell me a general way to solve the problems of this type where probability of $k$ consecutive success in $n$ trials is asked. In this question I was able to manually find the possible arrangements with consecutive successes but if the number of trials is high for example 200, how would I approach this problem then.
Thanks very much.
 A: I think your answer of 0.0648 is correct.
For a more general approach, consider the probability of not having 3 successes in a row out of $n$ trials; call this probability $P(n)$, and let's say the probability of a single success is $p$, with $q=1-p$. If we have $n$ trials, condition the probability on the number of successes at the end of the $n$ trials: the $n$ trials must end in $...F$, $...FS$, or $...FSS$, so we have the recursion
$$P(n) = q \; P(n-1) + pq \; P(n-2) + p^2q \; P(n-3) \qquad \text{for } n \ge 3$$
with $P(0) = P(1) = P(2) = 1$.
It's not hard to see how to extend this approach to longer strings of successes.
A: Let $X_i=\begin{cases}
    S, & \text{with prob 0.3 }  \\
    F, & \text{with prob 0.7 }  \\
  \end{cases}$
  for $i=1,2,3,4,5$  be IID for the outcome of the i-th day.
The probability you are asked is the following (if we are talking for exactly 3 days succes in a row)
P(Mr Li get home in time 3 days in a row)=
$ P(X_1=X_2=X_3=S)+P(X_1=F,X_2=X_3=X_4=S)+P(X_1=X_2=F,X_3=X_4=X_5=S)= 0.3^3+0.3^30.7+0.3^30.7^2=0.05913$ 
(which is the correct answer in the book)
