In the “Given Probability” video on PrepSwift, Greg does a problem two different ways.

**Way 1:**

**Way 2**

Then Greg does the following problem in way 2.

I tried to do this problem way 1.

Am I doing something wrong here? Why this discrepancy?

In the “Given Probability” video on PrepSwift, Greg does a problem two different ways.

**Way 1:**

Then Greg does the following problem in way 2.

I tried to do this problem way 1.

Am I doing something wrong here? Why this discrepancy?

Case that makes you happy are: HHHT, HHTH, HTHH, THHH

No. I am sorry. I couldn’t understand.

My question is about the rain problem.

With way 1, we get Probability = 1/33

With way 2, 1/11

Why this discrepancy?

**Given:**

Probability of rain on any given day = \frac{1}{3} so , probability of No rain = 1 - \frac{1}{3} = \frac{2}{3}

To Find: rain **at-least** twice, what is the probability of it rains on all four days.

Now, since it’s is at-least case , we need to consider the possibility of :

\text{P(rain at least twice) = P(2 successes) + P(3 successes) + P(4 successes)}\\
= 24/81 + 8/81 + 1/81 = 33/81

Now, what we want is rain on all 4 days , which is equal to:

\left(\frac{1}{3}\right)^4 = \frac{1}{81}

Now, we have our total cases(rain at-least twice) v/s our desired case(rain on 4 days):

\text{Probability} = \frac{\text{desired}}{\text{total}} = \dfrac{\frac{1}{81}}{\frac{33}{81}}=\frac{1}{33}

I am writing this for my own record.

I think the rain problem could not be solved using Way 1 because, unlike the coin problem in which the two outcomes are 50/50, in this problem, the two outcomes, rain and no rain, are not equally likely.

The following video explains something similar at the time stamp, although the reason for “balancing the tree” is not explained.

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