Okay. This post might be pretty long. Both of the following problems are from 5lb Chap 23, Probability and Combinatorics.

First check Question 34

**Picture 1**
Although, I solved this problem using a different and simpler approach, 5lb also suggested the following solution

**Picture 2**

Now see Question 36 from the same chapter

**Picture 3**
I calculated Quantity A using the *Inclusion-exclusion formula* and Quantity B by subtracting A from 1; however, 5lb answer key suggests another way to find Quantity B

**Picture 4**

Why didn’t we do the following, as we did in the solution of Q34? (please see picture 2)

Probability of no rain and no quiz = 1 - (0.3)(0.2)

For question 36 you have a total of 2 independent events. Giving you a total of four possible outcomes

Event |
Rain (0.3) |
No Rain (0.7) |

**Quiz (0.2)** |
0.06 |
0.14 |

**No Quiz (0.8)** |
0.24 |
0.56 |

For question 34 you only have 3 outcomes Jan being picked as the first kid \large\frac{1}{6}, Jan being picked as the second kid \large\frac{1}{6}, and Jan not chosen \large\frac{2}{3}.

Edit - In question 36 quantity A is “The probability that either or both events occur”, which is **0.06 + 0.14 + 0.24 = 0.44**

Your approach should work but it didn’t answer my question.

I think I’ve figured it out.

In solving Question 34 (see Picture 2), (1) we found out the probability of two events not occurring, (2) multiplied the two, and (3) then subtracted from 1. This gave us the probability of occurrence of event **at least once**.

In contrast, in finding Quantity B for question 36 (see Picture 4), we are simply finding the probability of two events not occurring. So, (1) we subtract both probabilities of occurrence from 1, and then (2) multiply them.