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?


Possible arrangements with three consecutive successes in 5 trials:


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.


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.


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)

  • $\begingroup$ This is actually incorrect because it does not take into account $P(X_1=X_3=X_4=X_5=S, X_2=F)$ $\endgroup$ – Couchy311 May 31 at 2:47
  • $\begingroup$ I agree with you technically, that's why I said "if we are talking for exactly 3 days S in a row"...it seems like the problem should have clarified that or give another answer in the back.. $\endgroup$ – sakas May 31 at 3:08
  • $\begingroup$ I see, thanks for clarifying :) $\endgroup$ – Couchy311 May 31 at 3:11

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