Probability Problem
Assuming events A, B and C are independent of one another, then your math is correct.
(These problems are examples of joint probabilities for independent discrete random variables.)
Kosmos
Global: @Calorie
MA Arcs in 4-star purgatory: Four in a Row (#2198) - Hostile Takeover (#69714) - Red Harvest (#268305)
I feel you pain.
On Saturday I was helping my niece with some maths assignments for about 2 hours.
Most of the time I can breeze through it as I use a lot of math on a daily basis, but I admit I had to look a few things up (eg congruency rules for an Iscosoles Rhombus).
@Catwhoorg "Rule of Three - Finale" Arc# 1984
@Mr Falkland Islands"A Nation Goes Rogue" Arc# 2369 "Toasters and Pop Tarts" Arc#116617
Right on. Many thanks for the confirmation.
I use to love math. Came very easy. Don't ask me to spell anything without a dictionary though
But wow, it's been a while. I guess a couple brain cells lived through them old college days. ha ha.
If she starts asking about Isosceles Rhombuses I'll give up and just take her for ice cream. Yipes.
Anyway, thanks again. Saved my reputation. Here's some for you.
-Morty
Paragon City Search And Rescue
The Mentor Project
Excuse me, but I was told that there would be no math.
I'm Tealeaf, and I approve of this message.
See [URL="http://boards.cityofheroes.com/showthread.php?p=2134817#post2134817"]Useful Tagging for the good of the Forum[/URL], my guide to tags.
There's always room for math.
Mini-guides: Force Field Defenders, Blasters, Market Self-Defense, Frankenslotting.
So you think you're a hero, huh.
@Boltcutter in game.
be careful not to step in the aftermath.
We don' need no stinkin' signatures!
Paragon City Search And Rescue
The Mentor Project
/em colors the powers on his second build while supersidekicked down and getting XP
Sorry, I was distracted there for a sec. Could you repeat your post?
Mini-guides: Force Field Defenders, Blasters, Market Self-Defense, Frankenslotting.
So you think you're a hero, huh.
@Boltcutter in game.
This is not quite on topic here, but I know a lot of you are VERY good at math, and quite a few at figuring out probabilities.
If someone could check my logic here, or even point me to where I can find out, I (and my pesky niece!) would be most grateful.
Probability problem
Given:
½ probability of A, = 0.5
1/3 probability of B = 0.3333 (Repeating)
¼ probability of C. = 0.25
Solve:
(1) A and B will both be successful? =
0.5 x 0.3333 = 0.167 16.7%
(2) A, B and C will all be successful? =
0.5 x 0.3333 x 0.25 = 0.042 4.2%
(3) None will be successful? =
(how I did this is reverse the probability for each set, then multiply. Not sure this is right)
0.5 x 0.6666 x 0.75 = 0.25 25%
It's been a long time since school, now my niece is asking me about this stuff... Please you math gurus, help me save the day!
-Morty