# Homework Help: Help with probability

1. Sep 16, 2010

### sneaky666

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})
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 whats the probability that A1 watches neither news?
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?
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) ?
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)?
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%

2. Sep 16, 2010

### Fredrik

Staff Emeritus
Re: help with probability!!

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).

3. Sep 16, 2010

### Redbelly98

Staff Emeritus
Re: help with probability!!

(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: Apr 25, 2017
4. Sep 16, 2010

### Redbelly98

Staff Emeritus
Re: help with probability!!

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.

I agree.
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.