An urn contains 3 blue and 7 red marbles. Marbles are drawn, one at a time, without replacement from the urn. Explain why the probability of that the third marble drawn is red, is equal to the probability of the first marble drawn is red.
Nice question requal
Consider 10 boxes in which you are going to put the marbles that you
select i.e one box for each marble.The first marble getting into the first
box,2nd marble getting into the second box and ...so on.
Now it is clear that the probability of red marble getting into any box is 3/10.
Hello, requal!
Another way of looking at the problem . . .
An urn contains 3 blue and 7 red marbles.
Marbles are drawn, one at a time, without replacement from the urn.
Explain why the probability of that the third marble drawn is red,
is equal to the probability of the first marble drawn is red.
Considering only the colors of the marbles,
. . they can be drawn in $\displaystyle {10\choose3,7} \:=\:120$ ways.
Imagine listing the 120 possible orders.
In how many of them is the third marble red?
It should be no surprise to us that: .$\displaystyle \begin{array}{c}\text{the third marble is red }\frac{7}{10}\text{ of the time} \\ \\[-4mm]
\text{the third marble is blue }\frac{3}{10}\text{ of the time}\end{array}$