Fax Multiple Conditional with airSlate SignNow

Fax multiple conditional

so let's build on this concept more by moving on from two events to a situation where three events are involved so I have three events here we'll call them a B and C so the question here is what is the conditional probability of a union B given a condition C in other words the possibility of a or B occurring in a set that contains all the C outcomes so if you look at this diagram here that would be this entire blue section which is events associated with a or B in my set C to represent this using a formula an easy way to do it is by starting with the formula for B a union B what is B a union B it is a probability of a plus the probability of B minus their intersection as we don't want any overlap or repeated terms showing up so the conditional probability of the Union will just be the conditional probability of a plus the conditional probability of B minus the conditional probability after intersection same formula with the condition attached to each term but there's one thing we don't quite know conditional probability of the intersection of two sets how do you find that remember conditional probability of a on B is nothing but a events common to B divided by all B events applying the same thing for a intersection B it should be a intersection B events common to C / all C events in other words my numerator will be the segment that contains events common to all these three a B and C so if you look at my diagram here the portion common to all the three events will be this blue portion a intersection B intersection C so my conditional probability of a intersection B now becomes B a intersection B intersection C divided by the probability of C which represents all these C events so now we have another problem how do we calculate the intersection of presets we've only dealt with two so far we'll use an example to understand this so here I have for red for black and for white balls placed in a box and I'm going to pick out three balls from it so the question to you is what is the probability that my first pick is a red ball the second pick is a black and the third is a white ball so let me draw my box here for red for blue and for green balls in it and the conditions I need to fulfill our first pick to be read second pick to be blue and the throat pick to be Queen what is the probability that all these three events are favorable we'll start with our first pick being read during my first pick I have four red balls for black and for green balls in this box say the probability of a red pick will be four over twelve four red balls twelve balls in total which is one over three so let's move on to calculating the conditional probability of the second draw being blue given that our first draw is red so if I'll take my second pick now I'm only left with three red balls in the boxes one red ball is already been taken out and then I have four blue and four green balls so the conditional probability of a blue pick is favorable outcomes divided by total outcomes which is four over eleven for blue balls out of eleven balls in total so moving on now what is the probability that my third pick is green given that two conditional probabilities are met the first condition is pick one is read and the second condition is pick two is blue so if I look at my box during the third row it has three red balls and three blue balls is as one of each were taken out during pick one and pick two and we have four green balls so conditional probability of a green ball being taken out in my third pick is for over ten four green balls out of ten balls in total so now if I want to find a probability that pick one is red pick two is blue and pick three is green in other words the probability that all these three outcomes are favorable I simply multiply the probability of the first draw being red with the conditional probabilities of my second and third draw being blue and green so this will be four over twelve into 4 / 11 into 4 / 10 here my 4 / 10 can be simplified as 2 / 5 + 4 / 12 will be 2 over 6 so when you multiply all of this you get 16 over 330 or point 0 2 or 2% so if you sum all this up here when we have three dependent events a B and C the intersection of all of this we'll be given by V of a into the conditional probability of B given a into the conditional probability of C given a and B but there's one more thing you might wonder about what would happen if my three sets are independent what would then be the probability that all of these are favourable so what we'll do is to start with the earlier formula of the intersection of a B and C for dependent events so here the conditional probability of B will be nothing but the probability of B as occurrence of a has no effect in B similarly the conditional probability of C will be nothing but the probability of C as the occurrence of either A or B doesn't really affect its likelihood then the probability of the intersection is given by P of a into P of B into B of C