I have simple probability problem.
I've researched (
Probability finding a card in $n$ draws? by @JMoravitz
and made my calculations. But I am not sure if the result is correct. Could you please assure me?
Context: a deck of 10 cards: 8-hearts and 2-diamonds. 10 people drawing 1 random card in turns. (The first draws from 10cards, the second – from 9cards, etc.) What is the probability of drawing a diamond?
Here is my calculations, visually outlined: (see pic)
Best Answer
Drawing without replacement, a symmetry argument like drhab's comment says that the probability the $n$th card is a diamond is $\frac{2}{10}$ ans the $n$th draw is equally likely to be any of the ten cards, and two of those are diamonds
If you want the longer calculation then you need to consider three cases:
No diamonds may have been chosen in the first $n-1$ draws. This has probability $\frac{8 \times 7 \times \cdots (9-n)}{10 \times 9 \times \cdots (11-n)} = \frac{(11-n) \times (10-n)}{10 \times 9 }$. If so, the conditional probability the next is a diamond is $\frac2{11-n}$
One diamond may have been chosen in the first $n-1$ draws. This has probability $(n-1) \frac{2 \times 8 \times 7 \times \cdots (10-n)}{10 \times 9 \times \cdots (11-n)} = \frac{2(n-1)(11-n)}{10 \times 9 }$. If so, the conditional probability the next is a diamond is $\frac1{11-n}$
Two diamonds may have been chosen in the first $n-1$ draws. This has probability $\frac{(n-1)(n-2)}{2} \frac{2 \times 1 \times 8 \times 7 \times \cdots (11-n)}{10 \times 9 \times \cdots (11-n)} = \frac{(n-1)(n-2)}{10 \times 9}$. If so, the conditional probability the next is a diamond is $\frac0{11-n}$
So the overall probability the $n$th card drawn is a diamond is $\frac{(11-n) \times (10-n)}{10 \times 9 }\frac2{11-n} + \frac{2(n-1)(11-n)}{10 \times 9 }\frac1{11-n}+\frac{(n-1)(n-2)}{10 \times 9} \frac0{11-n} = \frac{20-2n+2n-2+0}{90} = \frac{18}{90} = \frac{2}{10}$ as expected