Help with Probability: Steps and Solutions for Common Questions

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The discussion focuses on solving various probability problems, with users sharing their attempts and solutions. Key questions include calculating probabilities for specific events, such as the likelihood of watching news at certain times, the probability of at least one knee being sore, and the chance of drawing a specific card from a deck. Users express confusion about the correct application of probability rules, particularly regarding "or" scenarios and complementary events. The thread highlights the importance of clear step-by-step reasoning in probability calculations and the need for proper categorization of homework-related questions in forums.
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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.
 
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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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