Let's say you are given a bag with 3 identical marble inside. One of the marbles is white and the other 2 are black. If you can't see the marbles and you select one from the bag, what are the chances the marble is white?
Since only 1 out of 3 is white, the probability is 1/3.
Ok, now let's say I give you 3 chances to win. You get to select a marble 3 times. What are the chances that after 3 attempts of choosing a marble, at least one of the times you correctly select a white marble?
Well, if your chances are 1/3 each time wouldn't the answer be 1/3 for the first time, 1/3 for the second time and 1/3 for the third time? Add these chances up and you get 1/3+1/3+1/3 = 1. That is, after 3 tries you are guarenteed to choose a white marble even though each attempt is independent and random.
Our intuition does not serve us well in this case. The probability of 1 is incorrect. Let's take it step-by-step.
For our first attempt the chance of picking a white marble is 1/3. So far so good. Now for the second attempt, we need to consider the case in which we did not select a white marble in the first attempt. That is, 2/3 of the time in the first case we will not select a white marble. So given that we do not select a white marble in the first attempt, selecting a white marble in the second attempt is 2/3*1/3, which gives us 2/9.
Now for the third attempt, we need to consider the case in which our first and second attempts were not successful. 1/3 probability for the first attempt plus 2/9 probability for the second attempt gives us 5/9 probability we will be successful on the first and second attempts. Thus we are not successful on the first and second attempts 1-5/9 or 4/9 of the time. For the third attempt we have 1/3 chance of being successful given the first and second attempts fail. This gives us 1/3*4/9=4/27.
Now all we have to do is add these 3 numbers together: 1/3+2/9+4/27 = 19/27.
We have 19/27 or 70.4% chance of drawing at least one white marble from the bag after 3 attempts. Whew!
A simpler way is to think of the probability of NOT getting a white marble. 2/3 chance for the first attempt and 2/3 chance for the second attempt and 2/3 attempt for the third attempt. Since each attempt is independent, we can simply multiply the 3 probabilities together: (2/3)*(2/3)*(2/3)=8/27. We have a 8/27 probability of NOT getting a white marble. The probability of getting a white marble is 1 minus this probability or 1-8/27 giving us the same answer above.
Ok, but you may say why doesnt mulitplying 1/3 * 1/3 * 1/3 work? When we multiply 1/3 3 times, what we are really saying is: What is the probability I get a white marble the first time and get a white marble the second time and get a white marble the third time? The chances of getting 3 white marbles in a row is 1.3*1/3*1/3 = 1/27. We would expect this probability to be different and smaller than the probability of getting at least one white marble after 3 attempts.
Finally, you may ask how come adding 1/3+1/3+1/3 =1 does not work? This is a common but logical fallacy in reasoning most of us are guilty of at first glance. Adding together implies mutual exclusion, a subject to be covered later.
We can also obtain the answer by considering all of the possible scenarios and the probabilities of each scenario where B represents a black marble and W represents a white marble for an attempt. We can list all possible outcomes of the 3 attempts:
1. BBB = (2/3)(2/3)(2/3)=8/27
2. BBW = (2/3)(2/3)(1/3)=4/27
3. BWB = (2/3)(1/3)(2/3)=4/27
4. BWW = (2/3)(1/3)(1/3)=2/27
5. WBB = (1/3)(2/3)(2/3)=4/27
6. WBW = (1/3)(2/3)(1/3)=2/27
7. WWB = (1/3)(1/3)(2/3)=2/27
8. WWW = (1/3)(1/3)(1/3)=1/27
Now add up the probabilities for all of the scenarios in which you get at least one white marble, i.e. all of the scenarios except the first one: 4/27+4/27+2/27+4/27+2/27+2/27+1/27=19/27, which is the same as we obtained earlier.
Note how a seemingly simple problem can easily lead us astray. More on the way!
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