A box contains 3 yellow balls and 5 black balls. One by one

Let me offer another approaches:

1. There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that 2 of the balls are the same colour?

So there are five different balls: {1, 2, 3, 4, 5}

So we need the probability of {XXYZ}, for example {2, 2, 4, 5} in any order: \(P=\frac{C^1_5*C^2_4*}{5^4}*\frac{4!}{2!}=\frac{72}{125}\);

\(C^1_5\) - # of ways to choose which ball will be X (the ball which will be chosen twice);
\(C^2_4\) - # of ways to choose which balls will be Y and Z (the other 2 balls from 4 left);
\(5^4\) - total # of outcomes;
\(\frac{4!}{2!}\) - {XXYZ} can occur in several ways: XXYZ, ZYXX, XYXZ, ... basically the # of permutations of 4 letters out of which 2 are identical, which is 4!/2!.

2. A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

There is a shortcut solution for this problem:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Check similar problems:
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Also Probability and Combinatorics chapters of Math Book: math-probability-87244.html and math-combinatorics-87345.html

Hope it helps.