Does any one have "worked problems" or sample problems with answers in pdf for Conditional Probability , I really don't understand this concept ?
Thanks
Thanks
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Did you search the web? I easily found http://www.regentsprep.org/Regents/math/… which may help.
Here's how I look at conditional probabilities. You have this universe of events that may take place, with known probabilities. A conditional probability looks at some subset of those, and examines the probability within that subset.
E.g., the probability of rolling two dice and getting a nine is 4/36 = 1/9
That's because there are only 4 ways to get a 9: 36 63 45 54, out of 36 total ways to roll two dice.
Now let's restrict the situation. What's the probability of rolling a 9, given that one of the dice is a 3. Now we don't have the whole universe of 36 possilble rolls. We only have 11:
31 32 33 34 35 36 13 23 43 53 63. Only 2 are 9's: 36 63. So the conditional probability, "rolling a 9 given that one of the dice is a 3" is 2/9.
The formula for conditional probability just cut things down to these numbers. The formula would say that this = Pr(9 and a 3) / Pr(3).
Pr(total=9 and one die = 3) = 2/36 since there are 2 ways to do this out of the 36 total.
Pr(one die = 3) = 11/36
And again the conditional probability is 2/36 / 11/36 = 2/11.
Hope this helps. Your idea to get some practice is an excellent one. That's the best way to learn this well.
Here's how I look at conditional probabilities. You have this universe of events that may take place, with known probabilities. A conditional probability looks at some subset of those, and examines the probability within that subset.
E.g., the probability of rolling two dice and getting a nine is 4/36 = 1/9
That's because there are only 4 ways to get a 9: 36 63 45 54, out of 36 total ways to roll two dice.
Now let's restrict the situation. What's the probability of rolling a 9, given that one of the dice is a 3. Now we don't have the whole universe of 36 possilble rolls. We only have 11:
31 32 33 34 35 36 13 23 43 53 63. Only 2 are 9's: 36 63. So the conditional probability, "rolling a 9 given that one of the dice is a 3" is 2/9.
The formula for conditional probability just cut things down to these numbers. The formula would say that this = Pr(9 and a 3) / Pr(3).
Pr(total=9 and one die = 3) = 2/36 since there are 2 ways to do this out of the 36 total.
Pr(one die = 3) = 11/36
And again the conditional probability is 2/36 / 11/36 = 2/11.
Hope this helps. Your idea to get some practice is an excellent one. That's the best way to learn this well.