Results 1 to 3 of 3

Math Help - help with probability!!

  1. #1
    Senior Member
    Joined
    Sep 2009
    Posts
    299

    Exclamation 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%
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    692
    Hello, Sneaky!

    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) ?

    Sorry, your answer in incorrect.

    You started correctly, though . . .


    There are 4 Jacks: J\heartsuit,\,J\spadesuit,\, J\diamondsuit,\,\boxed{J\clubsuit}

    . . P(J) \:=\:\dfrac{4}{52}


    There are 13 Clubs: A\clubsuit,\,2\clubsuit,\,3\clubsuit,\,4\clubsuit,  \,5\clubsuit,\,6\clubsuit,\,7\clubsuit,\,8\clubsui  t,\,9\clubsuit,\,10\clubsuit,\,\boxed{J\clubsuit},  \,Q\clubsuit,\,K\clubsuit

    . . P(\clubsuit) \:=\:\dfrac{13}{52}


    \text{But }\,P(J\text{ or }\clubsuit) \;=\;\dfrac{16}{52}

    There are only 16 cards that will "win the bet".

    You counted the J\clubsuit twice.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Sep 2009
    Posts
    299
    ok i understand why its 16/52.
    This also means that the other 4 questions were correct?
    But now i have some other concerns.
    This is a question from my book.

    a guy has 3 cards
    1 red on both sides
    1 black on both sides
    1 red / black

    he puts 1 card on the table, and u only see the top side which is red.
    what are the chances of the other side being red
    my guess is 50/50, but the book says 2/3 using conditional probability.
    Why is it 2/3??

    Also a question about number 5
    Like u said in number 4, i counted the jack of clubs twice, so in question 5 when adding 55% and 30.95% to get chance of female or a long hair person, doesn't part of the long hair people already count in the chance of females?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  3. Joint Probability/Marginal Probability
    Posted in the Statistics Forum
    Replies: 0
    Last Post: March 2nd 2010, 10:15 PM
  4. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum