# Thread:

1. ## Question on conditional probability

I'm a bit stumped on this question. It seems simple but I don't know where to begin.

"There is a 50% chance of hard drive damage if a power line to which a computer is connected is hit during an electrical storm. There is a 50% chance that an electrical storm will occur on any given summer day in a given area. If there is a 0.1% chance that the line will be hit during a storm, what is the probability that the line will be hit and there will be hard drive damage during the next electrical storm?"

So it is the intersection between the line being hit and HDD damage. But how do you get there?

2. ## Re: Question on conditional probability

Originally Posted by showstopperx
I'm a bit stumped on this question. It seems simple but I don't know where to begin.

"Assume that there is a 50% chance of hard drive damage if a power line to which a computer is connected is hit during an electrical storm. There is a 5% chance that an electrical storm will occur on any given summer day in a given area. If there is a 0.1% chance that the line will be hit during a storm, what is the probability that the line will be hit and there will be hard drive damage during the next electrical storm in this area?"

So it is the intersection between the line being hit and HDD damage. But how do you get there?
Let's begin by defining events:

$A$ is the event that the hard drive is damaged.
$B$ is the event that the line is hit.
$C$ is the event that an electrical storm occurs.

$P(C) = 0.05$

$P(B|C) = 0.01$

$P(B|C)= \tfrac{P(B\cap C)}{P(C)} \implies 0.01 = \tfrac{P(B\cap C)}{0.05} \implies P(B\cap C)=0.0005$

$P\left[A|(B\cap C)\right] = 0.5$

$P\left[A|(B\cap C)\right] = \tfrac{P\left[A\cap (B\cap C)\right]}{P(B\cap C)} = \tfrac{P(A\cap B\cap C)}{P(B\cap C)} \implies 0.5 = \tfrac{P(A\cap B\cap C)}{0.0005}$ $\implies P(A\cap B\cap C)=0.00025$

$P\left[(A\cap B) | C\right] = \tfrac{P\left[A\cap (B\cap C)\right]}{P(C)} = \tfrac{P(A\cap B\cap C)}{P(C)} = \tfrac{0.00025}{0.05} = 0.005$

-Andy

3. ## Re: Question on conditional probability

Thank you sir!
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9601856470108032, "perplexity_flag": "middle"}