Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is ?

  • $\begingroup$ It's a geometric distribution. Do you have an expression for determining the mean of a random variable with a geometric distribution? $\endgroup$ – Brian Tung Feb 23 '17 at 22:35

First, think about it like this:

Since 40% of people are successful at getting a reserved train seat, 60% of people aren't. The 60 % of the people who were unsuccessful try again, and $0.6\cdot 0.6 = 0.36$, or 36% fail again. This continues, but there will always be a small percent of people who are unsuccessful at getting a train ticket, but those people don't matter.


The first time, 40% are successful, and the second time, $100-36 = 64$% are now successful, including the 40% who are successful the first time.

This means that the median person will succeed after 2 attempts.


The Mean is a little bit more tricky, but we can approximate.

  • The first time, $40$% were successful.

  • The second time, $64-40 = 24$% were successful.

  • The third time, after multiplying again, $21.6$% were unsuccessful, so $100-21.6$ = $78.4,$ and $78.4-64 = 14.4$% were successful.

You might notice a trend. $\frac {40}{24} = \frac {5}{3}$, and $\frac {24}{14.4} = \frac {5}{3}$. Assuming this continues, we can calculate the mean.

  • After counting the first three trips, the average is 1.312.

  • Counting only the first five trips, the average is (about) 1.9168.

  • After ten trips, the average is (about) 2.4244.

  • After 25 trips, the average is (about) 2.4999.

  • After 100 trips, the average is (about) 2.5.

We can say that the mean is 2.5.

I used Java Eclipse to calculate Mean. The code is here.

public static double Calc(int x) {
    double answer = 0;
    double power = 0.4;
    for (int i = 1; i < x + 1; i++) {
        answer += i * power;
        power = power * 0.6;
    return answer;
  • $\begingroup$ Although unrealistic, the problem asserts that there is a constant $40\%$ probability for success on any attempt. Don't make the problem out to be harder than it is. $\endgroup$ – Graham Kemp Feb 23 '17 at 23:23
  • $\begingroup$ Yes, but there is a $40$% chance until they are successful. So if you succeed at reserving a seat, you stop reserving one. $\endgroup$ – ETS1331 Feb 23 '17 at 23:38
  • $\begingroup$ Thank you. Suppose this question is asked and as you rightly said, the answer comes out to be 2.5. Do I keep it at 2.5 or round it off to 3 as the number of attempts must be an integer ? Basically, this is my doubt. I'm stuck between 2 options - whether to mark 2.5 as an answer or 3. It would be great if you can clarify this. @ETSPTQ $\endgroup$ – LumosMaxima Feb 24 '17 at 5:42
  • $\begingroup$ I think the average would be put at 2.5. For example, if you had the average of 4 and 5, you would get 4.5, instead of rounding to 5. $\endgroup$ – ETS1331 Feb 24 '17 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.