I have a bowl with 4 colored balls in it, yellow, red, green and blue.
I will take out all the balls one by one. Goal is to get the yellow ball.

1st attempt: I take one ball , write down the color and throw the ball away.
2nd attempt: 3 remaining balls, I take one, write down the color and trow it away.
3th attempt: 2 remaining balls. I take one, write down the color, trow it away.
4th attempt: 1 remaining ball left, take it, write down the color.

What is the probability I take the yellow ball in my first ór 3th attempt?

Thanks for the help.

I'm assuming you are randomly drawing the balls out?

If so, first attempt is 25%
Third attempt is 50%

what's 3th?

Yes I randomly take the balls out.
I know the first attempt gives me 25% and the 3th 50%. But I think you have to take into account the probability I take out the yellow ball at the 2nd attempt.

probability is 50%

25% on 1st, 25% on 2nd, 25% on 3rd, and 25% on 4th :-)

so probability of 1st or 3rd is 50%

On the 2ND attempt there are 3 balls in the bowl. How can P be 25%?

It's pronounced threeth

I agree

Can you explain your thoughts?

I was thinking something like this.

1st is 1/4 (4 balls in the bowl)
2nd is 1/3 (one ball is allready taken away
3th is 1/2 (two balls are allready taken away)

P1st (P-yellow * P-not yellow) = 1/4 * 3/4
P2nd (P-yellow * P-not yellow) = 1/3 * 2/3
P3th (P-yellow * P-not yellow) = 1/2 * 1/2

P1st or P3th is (1/4 * 3/4) + (1/3 * 2/3) + (1/2 * 1/2)

sure.
a few different ways to think about this. One would be intuitively there are 4 different balls all with equal probability in each case, so it has to be 1/4th for each ball on each of the 4 drawings.

Another way is using conditional probability, this will give the same results.

On the 1st drawing there are 4 balls in the bowl, so 1/4 chance the ball is yellow.
On the 2nd drawing there are 3 balls in the bowl. There is a 1 - P(1Y) = 1-1/4 = 3/4 chance one of those 3 balls is yellow. Therefore there is a 1/3*3/4 = 1/4 chance the ball is yellow.
On the 3rd drawing there are 2 balls in the bowl. There is a 1 - P(1Y) - P(2Y) + P(1Y)&P(2Y) = 1 - 1/4 - 1/4 + 0 = 1/2 chance one of those 2 balls is yellow. Therefore there is a 1/2*1/2 = 1/4 chance the ball is yellow.

Where P(1Y) is probability 1st ball is yellow. P(2Y) the 2nd ball. Also note that these event are mutually exclusive since the balls aren't replaced it is impossible for the yellow ball to be pulled both first and second, hence P(1Y)&P(2Y) = 0.

I feel like I'm in High School with this word problem.

Never had math in High School besides the first year. So I decided at age 38 to stop watching porn in my free time and study math instead.

Hope this ain't my midlife crisis.

nailed it

Okay thx I think I get it.
I am a noob, so I used a probability tree. Still got 50% so I am okay with the way I did it. ;)

There is one way yellow is taken on the first attempt.

Yellow at the 1st = 1/4


There are 6 ways yellow is taken on the 3th attempt.


1attempt 2attempt 3 attempt
1/4. 1/3. 1/2
Red Green Yellow
Red. Blue. Yellow
Blue. Green. Yellow
Blue. Red. Yellow
Green. Blue. Yellow
Green. Red. Yellow


1/4 + (6*(1/4*1/3*1/2)= 0,5

Thanks to make me rethink the solution.

Only one year of math in High School? wow. Well good on you for self improvement. Statistics and probability are my least favorite maths.

https://media.giphy.com/...3EWmJDBa5MjIr6g/giphy.gif

just the opposite for me - give me stats all day. The fun parts begins when you get to make your own questions and explore the correlation between questions.

damn, got here too late.

Probabilities are always interesting thought experiments.

Probability is awesome.

Here is a simple but fun puzzle.

You have 3 cards. One is black on both sides, one is red on both sides, one is black on one side and red on the other side.

You randomly draw one card out of a bag and see that it is red on one side. What is the probability that the other side is also red?

50%

1/3

0% correct answers so far :-P

75?

you're very wasteful, throwing your balls on the floor like this !

You fluffing assfluff! I was looking for my balls I keep them color coded because each one has a different drug in it, and you! You have to go and throw them like they are nothing because you got a math itch!

I learned that in first grade when I took the balls away from the other kids>

nice one :)