Help with Probability: Steps and Solutions for Common Questions

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Homework Help Overview

The discussion revolves around probability problems, including calculating probabilities related to events involving sets, news viewing habits, and health conditions. Participants are attempting to solve various probability questions and share their reasoning processes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to compute probabilities based on given scenarios, including events related to sets and conditional probabilities. Some participants question the reasoning behind the calculations, particularly regarding the interpretation of "or" in probability statements and the assumptions made about independence.

Discussion Status

Participants are actively engaging with the problems, providing their answers and reasoning. Some guidance has been offered regarding the interpretation of certain probability concepts, and there is a recognition of differing perspectives on the calculations presented. The discussion is ongoing, with multiple interpretations being explored.

Contextual Notes

There are indications that some problems may be textbook-style exercises, which could influence the expectations for the discussion. Participants are encouraged to clarify their assumptions and reasoning as they work through the problems.

sneaky666
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help with probability!

I need help with the following questions, i attempted to solve them, and i also put my answers here too.
When you are answering them, please show all your steps as i am kinda lost in this field.

1. suppose S = {1,2,3} and P({1,2}) = 1/3 and P({2,3}) = 2/3
compute P({1}), P({2}), P({3})
my answer
P({1}) + P({2}) + P({3}) = 1 <--- 1 being 100% and 0 being 0%
P({1}) + P({2}) = 1/3
P({2}) + P({3}) = 2/3
a) P({1}) + (1/3 - P({1})) + P({3}) = 1
P({3}) = 2/3
b) P({1}) + (2/3 - P({3})) + P({3}) = 1
P({1}) = 1/3
so then 1/3 + P({2}) + 2/3 = 1
so P({2}) = 0
therefore: P({1}) = 1/3, P({2}) = 0, P({2}) = 2/3

2. Suppose A1 watches the six o'clock news a 2/3 of the time and watches eleven o'clock news 1/2 of the time and watches both news 1/3 of the time. For a randomely selected day what is the probability that A1 watches only the six o'clock new?s, and what's the probability that A1 watches neither news?
my answer
chance for 6oclock = 2/3
chance for 11oclock news = 1/2 => complementary is 1/2
for only six o'clock news
has to watch the 6oclock news which is 2/3, and not watch the 11oclock news which is 1/2
2/3*1/2 = 1/3
so only watching 6 o'clock news is 1/3 chance...
for watching neither
avoid watching 6o'clock news which is 1/3 (complementary) and avoid watching eleven o'clock news 1/2 (complementary)
so its
1/3*1/2 = 1/6
so watching neither is 1/6 chance...

3. Suppose your right knee is sore 15% of the time and your left knee is sore 10% of the time. What is the largest possible percentage that at least 1 knee is sore? What is the smallest possible percentage that at least 1 knee is sore?
my answer
Largest
15%+10% = 25%
Smallest
0.15*0.10 = 0.015
so 1.5%

4. Suppose a card is randomely chosen from a standard 52 card deck.
What is the probability that the card is a jack or a club (or both) ?
my answer
chance of jack = 4/52
chance of club = 13/52
chances of club or jack = 4/52 + 13/52 = 17/52
chances of jack and club is 1/52

5. Suppose 55% of students are female and 45% are male.
44% of females have long hair and 15% of males have long hair.
What is the probability that a random student will either be female or have long hair (or both)?
my answer
chance of female: 55%
chance of long hair: (55%*44%) + (45%*15%) = 0.2420 + 0.0675 = 0.3095 = 30.95%
chance of female or long hair = 55% + 30.95% = 85.95%
chance of female with long hair: 0.2420 = 24.2%
 
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I just had a quick look at the last two, and I think you need to rethink the ones that contain the word "or". In the last one, how many males don't have long hair?

I will report this post to the moderators and request a move to the homework section. (Even if it isn't homework, it's still a textbook-style question).
 


(Moderator's note: thread moved from " Set Theory, Logic, Probability, Statistics")

Homework assignments or any textbook style exercises are to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned coursework or just independent study.
 
Last edited by a moderator:


sneaky666 said:
2. Suppose A1 watches the six o'clock news a 2/3 of the time and watches eleven o'clock news 1/2 of the time and watches both news 1/3 of the time. For a randomely selected day what is the probability that A1 watches only the six o'clock new?s, and what's the probability that A1 watches neither news?
my answer
chance for 6oclock = 2/3
chance for 11oclock news = 1/2 => complementary is 1/2
for only six o'clock news
has to watch the 6oclock news which is 2/3, and not watch the 11oclock news which is 1/2
2/3*1/2 = 1/3
so only watching 6 o'clock news is 1/3 chance...
for watching neither
avoid watching 6o'clock news which is 1/3 (complementary) and avoid watching eleven o'clock news 1/2 (complementary)
so its
1/3*1/2 = 1/6
so watching neither is 1/6 chance...
I'm not sure I agree with your reasoning process, but I do get the same answers: 1/3 to watch only 6 o'clock news, 1/6 to watch no news.

3. Suppose your right knee is sore 15% of the time and your left knee is sore 10% of the time. What is the largest possible percentage that at least 1 knee is sore? What is the smallest possible percentage that at least 1 knee is sore?
my answer
Largest
15%+10% = 25%
I agree.
Smallest
0.15*0.10 = 0.015
so 1.5%
No. The left knee is sore 10% of the time, so at least 1 knee is sore at least 10% (or more) of the time.
You seem to be calculating the probability that both knees are sore, assuming they are independent.
 

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