Do we need to use complement to solve the (rain, earth quake, flu) problem

Why wouldn't we add the probabilities such that p(r U e U f) = p(r) + p(e) + p(f)

Isn't that the union rule for independent events? It produces a probability of 0.95 which is incorrect but I don't understand why.

Thank you Warren Roberts

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I believe you're referring to this problem.

No, you're wrong. It should be $\text{ Probability of at least one of them happening = 100\% - Probability of none of them happening}$ as shown in the solution.

Why? Let's suppose you replace the numbers 30%, 20%, 45% in the problem statement with 70%, 80%, 90%, respectively. Will the answer turn out to be 70% + 80% + 90% = 240%? That's absurd!

But why is it wrong? Referring to the formula here, we have $P(H\cap E\cap F) = P(H)+P(E)+P(F)-P(H\cap E)-P(H\cap F)-P(E\cap F)+P(H\cap E\cap F).$

Since they are mutually independent events, we have $P(H\cap E) = P(H) \times P(E) , P(H\cap F) = P(H) \times P(F) , P(E \cap F) = P(E) \times P(F) , P(H \cap E \cap F) = P(H) \times P(E) \times P(F)$ .

So, without using complementary probability, the answer will still be the same: $P(H\cap E\cap F) = 30\% + 20\% + 45\% - 30\% \times 20\% - 30\% \times 45\% - 20\% \times 45\% + 30\% \times 20\% \times 45\% = 69.2\% \approx \boxed{69\%}$

Pi Han Goh - 2 years, 2 months ago

Tha really helps. Thank you

Warren Roberts - 2 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$