Does anybody know how to do probability stuff for MA 114. I have a test tomorrow and I need some help.

11/29/2007 7:23:03 PM

post questions

11/29/2007 7:25:24 PM

1. Tom and Alice work independently in an attempt to solve a certain problem, i.e. whether one of them solves it does not affect the chances that the other will solve it. The probability that Tom solves it is 0.2 and the probability that Alice solves it is 0.55.
What is the probability that the problem will be solved?


If the problem is solved, what is the probability that Tom solves it?


2. A student is trying to pass a competency exam. Each time she takes the exam she has a 20% chance of passing, and she is allowed a maximum of three attempts.
Draw a tree diagram to represent her attempts to pass the exam.


How many outcomes does your tree show?
What is the probability she will eventually pass the exam?
What is the probability she will take the exam three times?
What is the expected number of times she will take the exam?
What is the expected number of times she will fail the exam?

11/29/2007 7:46:23 PM

do you not know how to draw the venn diagram for (1) and the tree diagram for (2), or you don't know how to interpret them?

11/29/2007 8:13:44 PM

1. Let T = probability that Tom solves problem and A = probability that Alice solves problem. Then P(T)=.2 and P(A)=.55. You are trying to find the probability that the problem is solved so it can be Tom solving or Alice solving i.e. P(T or A). Then P(T or A) = P(T) + P(A) - P(T and A). Use the part of independence to figure out the rest.

11/29/2007 8:16:22 PM

^do they cover that kind of stuff in 114?

11/29/2007 8:18:12 PM

^ i would doubt it

11/29/2007 8:19:35 PM

how would they do it then?

11/29/2007 8:39:45 PM

Quote :

"venn diagram"

11/29/2007 8:40:23 PM

thanks for your help.

on the 2nd question i have figured out the first 3 parts but i can't figure out the last 2 parts to the question. And I figured it out by the tree diagram.

11/29/2007 8:54:41 PM

the expected value is just the average

11/29/2007 9:04:39 PM

just write it out like this:


attempts to pass: 1............... 2............................................................3
probability they pass: .2...(sum of all possibilities that require 2 attempts)....(" " 3 attempts)


then do 1*.2 + 2 * (P(x=2)) + 3*(P(x=3))

[Edited on November 29, 2007 at 10:50 PM. Reason : ...]

11/29/2007 10:49:47 PM