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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 ?

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  • $\begingroup$ 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
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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.

Median

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.

Mean

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;
}
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  • $\begingroup$ 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

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