In Alexander's program there are $50$ girls and $50$ boys. They use a cafeteria of 25 4-tables, enough to cover the peak hour.
Alexander would like to talk with girls rather than boys during the pause; so, he assumes on average a table will have $2$ girls and $2$ boys.
After taking his seat at a table, he discovers the cruel reality : on average there are only $1.5$ girls at his table and $2.5$ boys.
Why the girls are avoiding Alexander ? Why they are choosing other tables ?
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Alexander should consider that he's a boy himself, and that the expected value of a random variable (following a binomial distribution in this case) is a linear function. He should also consider working out and exercising proper hygiene.
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This question is very confusingly worded, and I don't entirely agree with its premise. But I believe the point is: Alexander is a boy, so any table he sits at will have one of the following distributions (representing by 'B' a boy and by 'G' a girl): BBBB, BBBG, BBGG, or BGGG (I've bolded Alexander in each case). This means the average number of boys is $\frac{4+3+2+1}{4} = 2.5$, and the average number of girls is $\frac{0+1+2+3}{4} = 1.5$.
This is a very basic question in probability theory. The fact that Alexander, who is a boy, certainly sits at his own table skewes the distribution -- for example, he will never see $4$ girls at his table.
The easiest way to think about it is that there are really $3$ places left after Alexander sits, for $50$ girls and $49$ (N.B. not $50$) boys. On average, a little over half of these seats ($3\times 50/99 \approx 1.515$) will be taken by girls, and a little less than half $(3 \times 49/99 \approx 1.485)$ will be taken by boys other than Alexander. So, Alexander is indeed slightly unlucky: he should see about $1.515$ girls and $2.485$ boys (including himself) at his table, but maybe he just did not get enough samples!