11:05 "So this is a 90% specific test." That is strictly not correct. Specificity refers to P(negative test given no cancer). You are talking about P(positive test given cancer) which is called sensitivity. In other words, a test that picks up 90% of cancers in a cancer group is said to be 90% sensitive (not 90% specific).
Probability of A (cancer) given dat the person has cancer = P(A|B) = 450/500 = 0.9. The person has cancer, but the test detects only 90% of persons with cancer. It is about the test and not about the cancer. 🤓
well imo he didnt define the events properly but heres what i think he meant to say event B is having cancer event A is having a positive test Probability of A given B well we know B is true, therefore that person has cancer, so the denominator is 500 the amount of people who tested positive and have cancer is 450 therefore p(A|B) = 450/500 = 0/.9
Yeah, it was defined poorly cause at first i was getting 0.8181818181. But if you define B as having cancer and A as having positive. Then the area of A & B is 450 divide that by the area of B which is 500, then you get 0.9.
We are calculating P(A|B) or the probability of P(A) (having a positive test) given P(B) (having cancer). So having cancer is given, we can are only concerned with the binary split at the top since this is the group that has cancer P(B). No we want to know the probability of a positive test in this group. We can see that 450 out of 500 had a positive test so we can conclude that the probability of a positive test given having cancer P(A|B) equals 0.9
So here's a problem: In the original Final Fantasy game, you had to build a party of 4 "warriors of light" (characters), and there were 6 classes each character could be, independent of one another: warrior, thief, monk, black mage, white mage and red mage. How many parties can the player build to start a new game with?
if order does not matter and with replacement (what is probably what you are looking for - warrior, monk or monk, warrior and one could choose same class, 2 warriors and 2 monks etc. ) nCr => 6! ( (6-4)! * 4!) = 15 different parties
@@frankg7048 15 corresponds to the number of parties you can create without repeting any class, and with order not mattering. This doesn't take any scenario in which repetitions occur into consideration.
Each probability is divided by the number of all occurences. Simplify in p(a&b) / p(b). You're left with #of occurrences of a&b divided by the # of occurrences of b .
no i think what he meant to say here that B referer to wathever probability you might think of regarding having or not cancer while A referer to wathever probability you might think of regardingthe test so test , make sense ?
I stopped watching after his formula gave a different value than P(A|B) = .5, which is the answer provided for the spades and black cards example. The formula would yield an answer of .25 Also, the random examples are given everywhere in textbooks. I will look for something with exercises and correct use of the conditional probability formula.
For these problems we use Hypergeometric Probability.
This was a Computational problem we had to resolve in card games like twenty years ago.
Great video Steve. Where to get a similar board?
the last 3 minutes was a bit rushed for me. looking forward to part 2.
P(A|C) Knowing C increase the probability of A. If C is not known the probability of A is 1/6 while if I know C happened the probability of A is 1/5
11:05 "So this is a 90% specific test." That is strictly not correct. Specificity refers to P(negative test given no cancer). You are talking about P(positive test given cancer) which is called sensitivity. In other words, a test that picks up 90% of cancers in a cancer group is said to be 90% sensitive (not 90% specific).
Can someone explain 10:55 on why it is 0.9
Probability of A (cancer) given dat the person has cancer = P(A|B) = 450/500 = 0.9. The person has cancer, but the test detects only 90% of persons with cancer. It is about the test and not about the cancer. 🤓
well imo he didnt define the events properly
but heres what i think he meant to say
event B is having cancer
event A is having a positive test
Probability of A given B
well we know B is true, therefore that person has cancer, so the denominator is 500
the amount of people who tested positive and have cancer is 450
therefore p(A|B) = 450/500 = 0/.9
Yeah, it was defined poorly cause at first i was getting 0.8181818181. But if you define B as having cancer and A as having positive. Then the area of A & B is 450 divide that by the area of B which is 500, then you get 0.9.
We are calculating P(A|B) or the probability of P(A) (having a positive test) given P(B) (having cancer). So having cancer is given, we can are only concerned with the binary split at the top since this is the group that has cancer P(B). No we want to know the probability of a positive test in this group. We can see that 450 out of 500 had a positive test so we can conclude that the probability of a positive test given having cancer P(A|B) equals 0.9
So here's a problem:
In the original Final Fantasy game, you had to build a party of 4 "warriors of light" (characters), and there were 6 classes each character could be, independent of one another: warrior, thief, monk, black mage, white mage and red mage.
How many parties can the player build to start a new game with?
if order does not matter and with replacement (what is probably what you are looking for - warrior, monk or monk, warrior and one could choose same class, 2 warriors and 2 monks etc. )
nCr => 6! ( (6-4)! * 4!) = 15 different parties
@frankg7048 Your assumption was correct, but unfortunetely the answer was not :/
@@saraiva407 why?
@@frankg7048 15 corresponds to the number of parties you can create without repeting any class, and with order not mattering. This doesn't take any scenario in which repetitions occur into consideration.
@@saraiva407 You are fully right. thx for correction. I took the wrong formula => it is (n+r-1) over r then => (6+4-1 over 4) => 9!((6-4)!*4!) => 7560
P(A|C) = 1/5
Agree
Dividing by probability is not obvious, should be explained better imho.
Each probability is divided by the number of all occurences. Simplify in p(a&b) / p(b). You're left with #of occurrences of a&b divided by the # of occurrences of b .
10:17 "Lets say that the test result being positive OR negative is event A" Everyone tested either positive OR negative, so p(A) = 1
am i crazy??? 0.55 is the probability that someone gets a positive test, which isnt how A was defined...
no i think what he meant to say here that B referer to wathever probability you might think of regarding having or not cancer while A referer to wathever probability you might think of regardingthe test so test , make sense ?
good
😁💪💪
I stopped watching after his formula gave a different value than P(A|B) = .5, which is the answer provided for the spades and black cards example. The formula would yield an answer of .25
Also, the random examples are given everywhere in textbooks. I will look for something with exercises and correct use of the conditional probability formula.
not the best video, kind of confusing