I picked one example from https://docplayer.net/6566428-Probability-exam-questions-with-solutions-by-henk-tijms-1.html
The problem is –
Bill and Mark take turns picking a ball at random from a bag containing four red balls and seven white balls. The balls are drawn out of the bag without replacement and Mark is the first person to start. What is the probability that Bill is the first person to pick a red ball
Where the solution is given already in that material, I failed to grasp it. Can you please help understand in intuitive way.
Any help will be highly appreciated.
Thanks,
Best Answer
The event $A_i$, the first red ball was taken out on the $i$th draw occurs when the first $i-1$ balls are white, then a red ball. So you need to choose the first $i-1$ white balls: ${7 \choose i-1}$, then choose the order in which they are placed (the permutation, $(i-1)!$), then pick the red ball to be the first ($4$ options) and then place the rest of the balls in any order. There are $3$ red balls to place and $7-(i-1)$ white ones, so the number of permutations is $(7-(i-1)+3)!$.
All this should be divided by the sample space, which is the permutations of all 11 balls, $11!$.
To find the probability asked for in the question, red should be on an even place so we only sum these probabilities for $i=2,4,6,8$.
This was the solution from the link. For me, it is much more intuitive to choose a sample space where the balls are the same. So the sample space is to choose the places of the red among the 11 possibilities: ${11 \choose 4}$ and the event $A_i$ occurs when there is a red ball in place $i$, so it is only left to choose $3$ places after it, i.e., among the other $7-(i-1)+3$ places. The probability is $$\tfrac{7-(i-1)+3 \choose 3}{11 \choose 4}$$