My maths teacher set this problem the other day: (Just clarifying that everyone knows how pool balls work, they are either stripey or solid colour) A bag contains one pool ball that is known to be a solid, a second ball is chosen in such a way that it has equal chance of being a solid or a stripe, this ball is then placed in the bag. A ball is now selected randomly from the bag and it turns out to be a solid. If somebody now selects a ball from the bag what is the probability of it being a stripe? I worked out an answer and i was told that it was wrong. I will post my answer shortly so as not to influence your decisions. Please could people have a go at this and post their reasoning.
There is a 25% chance, I think... There is a 50% chance the ball will be a stripe, and a 50% chance you will choose that ball.
Call the balls A (solid) and B (50/50 solid/stripe). The second draw has a 50/50 chance of being A or B. So the chance that the second draw is solid is: 50% of being A x 100% that A is solid + 50% of being B x 50% of B solid = 75% solid therefore: 75% solid 25% stripe
It is complicated because you have to think about the fact that it is more likely that the ball added was a solid than a stripe because of the fact that a solid ball was pulled out
I realise that my initial logic was wrong I worked on the basis that after the second ball was added there was a 50% chance of there being 2 solid balls and a 50% chance of there being one of each type. After a solid ball is now removed you can subtract a solid ball from each situation leaving a 50-50 chance I didnt consider the fact that the fact that the solid ball was remove before indicates a higher probability of selecting a solid ball.
Actually, this is a good problem. Very interesting. I think the probability is 1/3. There are two equal possibilities. Two solids and one solid and one stripe. So if I pull out a solid, only one of the three is paired with a stripe. So the probability is 1/3.
I can never remember the applicable formulae or the text book logic for such problems. I usually getr the correct answer by imagining it as a sampling problem. Suppose you did this process 60 times (chosen due to being divisible by a lot of different factors). 30 times a solid ball was transferred, and 30 times you drew a solid ball from the bag. 30 times a striped ball was transferred. In 15 of the 30 cases, you expect to draw a solid ball the second time. In 15 cases, you expect to draw a striped ball. From the above, a solid ball was drawn a total of 45 times: In 15 cases, a second draw would result in a striped ball. In 30 cases, it would result in a solid ball. The above leads me to believe that the probablity of drawing a striped ball is 1/3.
1/4 theres only 1 way in which the 2nd one can be a stripe: the 1st ball drawn has to be the one which ws already in it (the prob of that bein 1/2) and the ball that was placed was a stripe (the prob of that bein 1/2). hence prob of drawin the stripe after a solid = 1/2 * 1/2 =1/4
Call the solid ball originally in the bag #1, and the one you added #2. Given that you picked a solid ball, there are 3 (equally likely) possibilities: • #1 is solid, #2 is solid, and you picked #1. • #1 is solid, #2 is solid, and you picked #2. • #1 is solid, #2 is striped, and you picked #1. ... so the probability that the second one picked is striped is 1/3.
The problem is, in none of those situations did you pick a ball that was striped. The fourth possibility is: #1 is solid, #2 is striped, and you picked #2. Therefore, there is a 1/4 chance you will choose the striped ball.
you have a 50/50 chance of selecting a striped ball the first ball is NOT a random choice, we already know there is a solid and you picked it. so basicly you are asking "what is the chance of selecting a solid, given the selection is equal for solid and stripe"
The probability that the remaining ball is a stripe is 1/3. There are three equally-likely ways in which one can draw a solid ball, only one of which corresponds to there being a stripe left in the bag.
So far, I have yet to see an explanation as correct and as easy to undrstand as my analysis. The probability of picking a striped ball is 1/3. Any other answer indicates a faulty analysis. I am not sure that those giving 1/3 as the answer have convincing arguments supporting their view. I will not bother to repeat my previously posted analysis.
How can those all be equally likely, you have two options which involve a solid being added and only 1 option that involves a stripe being added
i don't know how to figure percentages but here is the scenario: on each attempt you will either have a 50% chance of drawing a stripe or a 0% chance. if we label the balls as A and B and the A ball is the known solid: you draw A then you have a 50% chance of getting a stripe on the next draw. you draw B then you have 0% chance of drawing a stripe on the next draw.
