I have to take an engineering statistics course this summer and as someone who has not liked statistics much in the past, your channel and videos are a lifesaver!
Hi Jeremy. Thank you so much for the video! I have a quick question regarding the sample spaces of the partition and the even A. For example, in the second exercise (randomly selecting a ball from a cup), the sample space of the first event (selecting a cup) is {cup 1, cup 2, cup 3, cup 4}, however, the sample space of the second event (selecting a ball) is {blue, red}. Since the sample spaces are different, could you please elaborate on why would be the law of total probability works in the case? Thanks in advance.
The probability experiment is randomly selecting an urn, then randomly selecting a ball from that urn. There are different ways we could define the sample space, but one is S = {1B, 1R, 2B, 2R, 3B, 3R, 4B, 4R}. It's the same idea as if we have a probability experiment where we toss a coin twice and observe whether heads or tails comes up on each toss. There are 4 possibilities, S = {HH, HT, TH, TT} (where HT represents getting heads on the first toss and tails on the second). A: The first toss is heads, and B: the second toss is heads, are defined on the same sample space.
For the final problem (where all the balls are mixed), I got P(Ball = Blue) = 0.382, and I can't figure out why the probability would be higher if all the balls are mixed up, compared with when they're separated into four urns?
What I think is that since drawing a ball from the urn's is weighted, meaning that each urn has a different composition of blue balls, so drawing a blue ball in each scenario would be different, still it is 37.7% likely to draw a blue ball. But, mixing all the balls into one huge bag breaks the idea of weights as everything is in one single place, therefore drawing a ball from a huge tank of balls would make the probability 13/34. Anybody can add in or correct me if I am wrong. Thanks.
@@pavanchaudhari5245 I also think it is because of weighting. I think it is like comparing average vs weighted average. If you do the same in the machine exercise, outcomes will be very different I think. If instead of 5/9 in the last urn it would be 5/50, calculating unweighted would skew the probability.
I think if you put all the balls together, then each blue ball will have equal chance to be picked. But if you put the balls in urns, then choose a ball is first depends on which urn do you pick. Use this extreme example, if I rearrange the balls. I put all the red balls in one urn, and put 4, 4 and 5 blue balls in the other 3 urns. there are 3/4 chance to select the urns with only blue balls, and the probability to pick a blue ball is 0.75 now. 1/4 x 1 + 1/4 x 1 + 1/4 x 1 + 1/4 x 0 = 3/4 = 0.75
Hi Jeremy. Thank you so much for the video! Can you please tell me that how identify when we have to use this particular "Law of Total probability" for which type of question
Thanks for the nice video. Please give us some examples where events are not mutually exclusive and non exhaustive. Will be looking forward to your video on practical applications with Baye's theorem.
Sir I have a qsn. We are using conditional probability because the events are dependent. But in the case of independent events I think the law of total probability will be like multiplication of individual events instead of conditional probability for 2nd event. Am I right???
I don't understand what you're asking . B_1 through B_k are mutually exclusive events that cover the sample space. A is another event in that sample space. A is going to intersect with at least one of the Bs. Why is the existence of A in question?
Imagine an extreme scenario: 1000 balls, with 1 blue and 999 red. Put them all in a single urn, and randomly pick a ball. The probability you pick a blue ball is 0.001. Now put the 999 red in an urn, the 1 blue in another urn, and choose between the urns with probability 0.5 then pick a ball from that urn. What's the probability you get a blue ball? 0.5. If each of the two urns contained 500 balls (the balls were evenly split), then the probability of getting a blue ball would be 0.001. The different number of balls in each urn messes with this.
I'm not sure at what level you're asking this question. Events are mutually exclusive if they share no portion of the sample space. In the Venn (Euler, actually) diagram examples, they were mutually exclusive if they didn't overlap (didn't share any sample points -- any portion of the sample space). I also showed a situation in which events shared common ground, and said they were not ME. In the example with the machines, each part was made by one and only one machine.
"One of these urns is randomly selected, in such a fashion as each urn is equally like to be chosen." There are 4 urns, and they all have the same probability of being selected.
We're asked for the probability we draw a blue ball, if we select one of the 4 urns at random and then draw a ball. P(Blue) = sum P(urn_i)P(blue | urn_i) = sum (1/4)*proportion of blue balls in that urn. I don't see how 1/number of blue balls could come into play.
Great video! my answer for the quizz : B = "Picking a blue ball" NB : One of the urn has 13 blue balls and 21 red balls and the other urns are empty... P(B)= 1/4 * 13/34 = 0,096 do we all agree? if not, tell me why in the comment section? thanks!
@@Loona_r_ B = B n U1 + B n U2 + B n U3 + B n U3 P(B) = P(B n U1) + 0 + 0 + 0 P (B) = P(U1) P(B/U1) P(B) = 1/4 * 13/34 It is possible because the URNs are mutually exclusive and exhaustive events...same as the balls
