The Q2 in the given probability prepswift video is:
The probability of rain on any given day is 1/3. In a four-day period, given that it rains at least twice, what is the probability that it rains on all four days?
I tried to solve it using both the approaches mentioned in the video, the case approach and the probability approach
for the case approach, I get 1/11 since we effectively get 3 cases:
Case 1: RRRR = 1
Case 2: RRRN = 4
Case 3: RRNN = 6
Total =11. Our case = 1. Hence 1/11.
Using the probability approach, I am able to get the correct answer as in the video, which is 1/33. Can anyone help understand where I am going wrong in the case approach?
The probabilities of the three cases aren’t the same though.
Sorry but this was not mentioned in the video. It seemed we had to derive this kind of reasoning.
Also, why does it work in cases where the probability is equal and does not work in case where the probability is not equal? As a general rule, can we state that where probabilities are equal for all cases we can use either method (case or probability) and for unequal probability, the case method does not work?
I am not sure that the video implies that you could always do both methods. I think the probability method is mainly discussed on the video.
Because in the case approach, you are assuming that the probability of a RRRR case is the same as the probability of RRRN. When the probabilities are equal (i.e. 0.5), this is fine because the probability of a RRRR and RRRN case is the same, that is, 1/16. This is not the case when the probabilities are different.
Yes… but I would recommend sticking to the probability method unless you know what you are doing.
Hi, returning to this months later because I had the same question / observation. The video never mentions that using the cases approach (as you showed in your work) only works when the probabilities of both events are =. Both in the video description and the video itself, practice Qs using this approach are solved with no mention that this approach will not otherwise work.
One thing I noticed when I solved like you and got 1/11 is that if you multiply 1/11 by the probability of rain (1/3) you get the correct answer 1/33. I guess this makes sense because in this case, we have 1/11 cases where it rains on all four days. The probability that it rains on any given day is 1/3, so to get the probability that it rains on all four days, we multiply the one case we like (1/11) by the probability of rain. At least, that is my reasoning. I would be interesting to try on other practice Qs.
It just looks like you’re reverse engineering the answer through context because as it stands your math doesn’t make much sense to me.
You can use the cases approach; you just have to deal with the corresponding probabilities as well alongside the number of arrangements.
\frac{1\cdot \left(\frac{1}{3}\right)^4}{1 \cdot \left(\frac{1}{3}\right)^4 + 4 \cdot \left(\frac{1}{3}\right)^3 \left(\frac 23\right) + 6 \cdot \left(\frac 13\right)^2 \left(\frac 23\right)^2} = \frac{\frac 19}{\frac 19 + \frac 89 + \frac {24}9} = \frac{1}{33}
Notice how the coefficients are the number of corresponding arrangements, but we also “factor” in the probabilities of each such arrangement.
Okay that makes sense thank you so much!