The fourth option is the stripe added, and the stripe removed first and the solid removed second. The problem asks for the probability that the second ball is a stripe, given that this fourth possibility did not occur. Here's the argument again, stated a little differently (S = Solid, $ = Stripe): The bag either contains (S, S) or (S, $), with each case having a <sup>1</sup>/<sub>2</sub> chance of occuring. If the bag contains (S, S), the first and second balls picked will both be S, so the probability of picking (S then S) is <sup>1</sup>/<sub>2</sub> * 1 = <sup>1</sup>/<sub>2</sub>. If the bag contains (S, $), there's a <sup>1</sup>/<sub>2</sub> chance that the first one picked is S (and the second is $), so the probability of picking (S then $) is <sup>1</sup>/<sub>2</sub> * <sup>1</sup>/<sub>2</sub> = <sup>1</sup>/<sub>4</sub>. So the probability of each possibility is: P(S then S) = <sup>1</sup>/<sub>2</sub> P(S then $) = <sup>1</sup>/<sub>4</sub> P($ then S) = <sup>1</sup>/<sub>4</sub> We're given that the first ball picked was S, so this rules out the last possibility on the list ($ then S), so: P(second is $ | first is S) = P(S then $) / (P(S then $) + P(S then S)) = <sup>1</sup>/<sub>4</sub> / (<sup>1</sup>/<sub>4</sub> + <sup>1</sup>/<sub>2</sub>) = <sup>1</sup>/<sub>3</sub>;
i believe you are in error if the bag contains 2 solids then the probability that you draw 2 solids is 1
It's no error, you cut off the quote at an illuminating bit. The probability there are two solids in the bag is 1/2. In this case the probability of drawing S then S is 1, thus his calculation (1/2)*1=1/2 for the overall probability of drawing S then S.
It doesn't state that when you place the randomly selected ball into the bag that it's done without seeing what it actually is, therefore you could suggest that you would know the outcome based upon what balls were put into the bag in the beginning. Namely if you know that you put solids in the bag, then you know there is no chance it will be a stripe, if you put a solid and a stripe in the bag you'll know that it will be a stripe.
You could also suggest there's not enough information to solve the problem. I mean it says you select "a ball" from the bag. I need to know how many soccer balls are in the bag, they aren't striped. Also, it doesn't stipulate that this bag isn't see through. Or that 'bag' isn't short for 'bag lady'. In this case, I'm not the 'somebody' that is reaching inside pulling the balls out.
I did it with a truth table: <table border="1" cellspacing="0" cellpadding="4"><tr><td colspan="4"><p align="center">Second ball added is solid</td><td width="63" style="border-top: none; border-bottom: none;" rowspan="5"> </td><td colspan="4"><p align="center">Second ball added is stripe</td></tr><tr><td colspan=2 rowspan=2> </td><td colspan="2" valign="top"><p align="center">1<sup>st</sup> draw</p></td><td colspan=2 rowspan=2> </td><td colspan="2" valign="top"><p align="center">1<sup>st</sup> draw</p></td></tr><tr><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td></tr><tr><td rowspan="2" valign="center">2<sup>nd</sup> draw</td><td><p align="center">Ball 1</p></td><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFFFA0">Solid,Solid</td><td rowspan="2" valign="center">2<sup>nd</sup> draw</td><td><p align="center">Ball 1</p></td><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #A0A0A0">Stripe,Solid</td></tr><tr><td><p align="center">Ball 2</p></td><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFFFA0">Solid,Solid</td><td><p align="center">Ball 2</p></td><td style="background: #FFA0A0">Solid,Stripe</td><td style="background: #A0A0A0">Stripe,Stripe</td></tr></table> The grey cells are eliminated by the first draw, leaving six possibilities if the first ball drawn is solid. The red cell is the only one of the six with the second ball being a stripe; a one in six chance. If the first ball drawn is not returned to the bag, then the truth table looks like this instead: <table border="1" cellspacing="0" cellpadding="4"><tr><td colspan="2"><p align="center">Second ball added is solid</td><td width="63" style="border-top: none; border-bottom: none;" rowspan="5"> </td><td colspan="2"><p align="center">Second ball added is stripe</td></tr><tr><td colspan="2" valign="top"><p align="center">Ball drawn 1<sup>st</sup></p></td><td colspan="2" valign="top"><p align="center">Ball drawn 1<sup>st</sup></p></td></tr><tr><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td></tr><tr><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFA0A0">Solid,Stripe</td><td style="background: #A0A0A0">Stripe,Solid</td></tr></table> And the answer is one in three.
i don't see how you guys are coming up with this 1/3 bit it is stated in the problem that the first ball is a solid. if you drew the known solid then you have a 50% chance of drawing a stripe. if you drew the random solid then you have a 0% chance of drawing a stripe. is my reasoning correct? like i said earlier i don't know how to figure percentages and that is probably what's throwing me.
Hi leopold, It's more likely that the first ball drawn was the known solid, simply because we know it was actually solid. The maths goes like this: The probability that some statement A is true, if it is given that another statement B is true, is: P(A | B) = P(A and B)/P(B) Reads as: The probability of A given B is the probability of A and B over the probability of B. More fully, the probability that A is/will be true if it is given that B is true is the probability that both A and B are/will be true divided by the probability that B had of being true before it was given. Here, A is "The first ball drawn was the first ball added", and B is "the first ball drawn was solid". P(B) = 0.75 (I've skipped a step here, but I think it's an easy one). P(A) = 0.5 P(A and B) is not known, but we can calculate it using the same rule as before with A and B switched, because we know that P(B|A) is 1: P(B | A) = P(A and B)/P(A) 1 = P(A and B)/0.5 P(A and B) = 0.5 P(A | B) = P(A and B)/P(B) = 0.5 / 0.75 = 2/3 So the chance that the first ball drawn was the known solid is 2/3.
The initial probabilities for the balls in the bag are 75% solid, 25% striped. This can be represented by 3 solid and 1 striped ball in the bag. Now take half of the balls out of the bag and assume they are both solid. This leaves 1 solid and 1 striped ball in the bag, so the probability of drawing a striped ball next is 50%. Thomas
What you've described is not at all the same problem. Here, learn about conditional probabilities: http://mathworld.wolfram.com/ConditionalProbability.html http://en.wikipedia.org/wiki/Conditional_probability lots more in google.
Forget about the links. They are not relevant and appropriate here. The solution I suggested <i>is</i> the same problem, just scaled up by a factor 2, which leaves the probabilities unchanged (whether you have 1/1 solid/striped balls in the bag or 0.5/0.5 doesn't matter; you could also have 100/100, the probability would still be 50%). You can also use the following argument: if the second ball added to the bag is a striped one, the probability of drawing it is 100% if the ball first drawn is a solid one; if the second ball added is a solid one, the probability of drawing a striped is 0% as there are no striped balls in the bag at all; the average of these two possibilities is 50%. I would recommend however in these kind of cases to always turn the probabilities into an integer number of objects (like in my first solution above); this makes the solution more intuitive and you avoid the pitfalls of working with formal probabilities alone; you just have to take care that you scale up all numbers by the same factor (e.g. when drawing balls like in this case). Thomas
If you scale the problem up by a factor of 2 (not a well defined operation), it becomes: "If a bag contains 4 balls, 2 of which are solid, and 2 of which are either solid or striped (50% chance either way), what is the probability that the third and fourth balls picked are striped, given that the first two picked are solid?" Generally if you double all the numbers in a probability problem, the resulting probabilities don't stay the same. The probability of picking 1 red ball from a bag containing 1 red and 1 blue ball is 1/2. The probability of picking 2 red balls from a bag containing 2 red and 2 blue balls is 1/4 (assuming the first ball picked is replaced before the second draw), or 1/6 (assuming no replacement). Careful with this approach - you're twice as likely to pick (on the first draw) a solid ball from a bag containing two solid balls than if the bag contains one solid and one stiped ball, so the average should be weighed accordingly. Suppose the bag contains 1 million balls, and you know that either they are all solid, or one is solid and the other 999,999 balls are striped (there's a 50% chance either way). If one ball is then picked and turns out to be solid, is the probability that the second one picked is striped still 1 in 2? Are the two possibilities: • "you picked one of the million solid balls in the bag on your first draw" and • "you picked the only solid ball out of a bag containing 1 million balls on your first draw" equally likely? Dinosaur did this on the first page of this thread:
50% obviously, as the two remaining balls are either both solid or both striped. However, you are confusing matters again by using probabilities. I have suggested putting 3 solid and 1 striped ball in the bag exactly to avoid this. You can also put 300 solid and 100 striped balls in the bag, which also corresponds to the given initial probabilities. If you now take 200 balls out and assume they are all solid (which is equivalent to taking 1 solid ball out of a total of 2), you are left with 100 solid and 100 striped balls, so there is a 50/50 chance of drawing a striped ball at the next attempt. This assumes statistically independent draws. If the condition is that 2 red balls are being picked together, the probability is still 1/2. The first draw is not an issue. It has already been made and it is a solid ball with 100% certainty. There is only 1 ball left in the bag and this can only depend on the probability for the second ball added (because you know also with 100% certainty that the first ball added was a solid ball). From what I said above one has to conclude that the probability is indeed 1/2 here as well. Again, the details for the first draw don't matter. The only relevant thing is that one solid ball was with certainty in the bag, and that a solid ball was drawn first. Thomas
You've seen conditional probability before then? You should understand how it applies to this problem, and how it was used above by Pete and pryzk. No it's not the same problem at all. You're trying to combine the two initial setups solid/solid and solid/stripe into one bag, then removing a solid from each. You are ignoring the fact that since the first draw was a solid, you are more likely to be in the solid/solid case, yet you are giving them the same weight. We have more information after the first draw. Those two events are not equally likely given the fact that you had already drawn a solid ball. I would recomend you get a friend and some pool balls and do a hundred trials and see what happens. You're serious? The 1/1,000,000 probability of drawing the lone solid ball from the bag with the striped balls doesn't make you suspect you aren't in this case? Have you taken any statistics at all? I have some great games for us to play. Bring lots of money.
Apply Bayes' Theorem, => 1/3. P(2nd is Striped | 1st is solid) = P(1st is solid | 2nd is striped) * P(2nd is striped) / P(1st is solid) = (1 * 1/4 ) / (3/4) = 1/3
You should read the task properly first before making any comments: there is only a 50% chance of the solid ball being a lone ball. The solid ball can in fact be completely neglected here. You know you put a solid ball in and you know you drew a solid ball out, so you can completely take it out of the equation. What's left is only the probability of the second ball added to the bag being solid/striped, and this is 50/50 according to the task. Anyway, the 50% result follows actually also from Pete's 'truth table': <table border="1" cellspacing="0" cellpadding="4"><tr><td colspan="2"><p align="center">Second ball added is solid</td><td width="63" style="border-top: none; border-bottom: none;" rowspan="5"> </td><td colspan="2"><p align="center">Second ball added is stripe</td></tr><tr><td colspan="2" valign="top"><p align="center">Ball drawn 1<sup>st</sup></p></td><td colspan="2" valign="top"><p align="center">Ball drawn 1<sup>st</sup></p></td></tr><tr><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td></tr><tr><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFA0A0">Solid,Stripe</td><td style="background: #A0A0A0">Stripe,Solid</td></tr></table> The left table gives 0% probability of drawing a striped ball, the right table a 100% probability, so the average of both is 50% (you have to average the two as the two cases for a solid/striped second ball added are mutually exclusive; since the two cases are equally likely, the average obviously must be unweighted). Thomas
So you don't know what a conditional probability is then? There's nothing wrong with that, except when you dismiss it as being irrelevant without knowing what it is. We have the two bags in front of us, 1,000,000 balls in each, one all solid one 999,999 stripe and one solid, bags are closed and we can't tell which is which. We flip a coin and pick one bag to pull balls out of. We pull out a solid ball. Here's the wager. If the next ball pulled out of this bag is striped, I will give you $1000, if it is solid, you give me $1. Willing to play?
They are NOT equally likely. Before you draw anything, each column has is equally likely, i.e. 1/4 chance of occuring. After the first draw, we are given more information, we know we are in the first 3 columns. Each *column* is still equally likely, i.e. we have a 1/3 of being in each. It's equivalent to just drawing 2 balls a bunch of times and throwing away any of the draws that give a stripe first, what proportion of the remaining draws would have a stripe in them? Seruiously, get some balls and try it.
