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Help with probability

  1. Sep 16, 2010 #1
    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 whats 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%
     
  2. jcsd
  3. Sep 16, 2010 #2

    Fredrik

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    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).
     
  4. Sep 16, 2010 #3

    Redbelly98

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    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
  5. Sep 16, 2010 #4

    Redbelly98

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    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.
     
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