Results 1 to 4 of 4

Math Help - Further Probability problems

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    105

    Further Probability problems

    1)A bag contains five white, six red and seven blue balls. If three balls are selected at random, without replacement, find the probability they are:

    -All different colors


    2)A four digit number(with no repetitions) is to be formed from the set of numbers (0,1,2,3,4,5,6,7). Find the probability that the number:
    a)is even
    b)is odd
    c)is less than 4000
    d)is less than 4000, given that it is greater than 3000
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    1) \displaystyle 18 balls in total. So the probability of getting all different colours would be:

    \displaystyle \frac{5}{18} \times \frac{6}{17} \times \frac{7}{16} = \frac{35}{816}.


    2)
    a) You require the last digit to be even. So the probability is \displaystyle \frac{4}{8} = \frac{1}{2}.

    b) You require the last digit to be odd. So the probability is \displaystyle \frac{4}{8} = \frac{1}{2}.

    c) You require the first digit to be a 0, 1, 2, 3. So the probability is \displaystyle \frac{4}{8} = \frac{1}{2}.

    d) There are \displaystyle {8\choose{4}} = 70 different combinations of 4 numbers from the 8 given.

    It has to be greater than 3000 and less than 4000. This means you know the first digit has to be a 3 and there are 7 digits left to choose from.

    That means you will be finding the number of combinations of 3 digits from the 7 remaining.

    There are \displaystyle {7\choose{3}} = 35 combinations.

    That means the probability of having a number less than 4000 and greater than 3000 is \displaystyle \frac{35}{70} = \frac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, johnsy123!

    \text{1)A bag contains five white, six red and seven blue balls.}
    \text{If three balls are selected at random, without replacement,}
    \text{find the probability they are all different colors.}

    There are: . {18\choose3} \:=\:816 possible outcomes.

    There are: . {5\choose1}{6\choose1}{7\choose1} \:=\:5\cdot6\cdot7 \:=\:210 ways to get one of each color.

    Therefore: . P(\text{all different colors}) \;=\;\dfrac{210}{816} \;=\;\dfrac{35}{136}





    \text{2) A four-digit number (no repetitions) is to be formed}
    \text{from the set of numbers (0,1,2,3,4,5,6,7).}

    Assuming that the number may not have a leading zero,
    . . there are: . 7\cdot7\cdot6\cdot5 \:=\:1470 possible 4-digit numbers.




    \text{(a) Find the probability that the number is even.}

    There are two cases to consider:
    . . The last digit is 0.
    . . The last digit is 2, 4 or 6.

    The last digit is 0: . \square\:\square\;\square\;0
    . . The other 3 spaces can be filled in: . 7\cdot6\cdot5 \:=\:210 ways.

    The last digit is 2, 4 or 6: .3 choices.
    The first must not be 0: .6 choices.
    The other 2 spaces can be filled in 6\cdot5\,=\,30 ways.
    . . There are: . 3\cdot6\cdot30 \:=\:540 ways.

    Hence, there are: . 210 + 540 \:=\:750 even numbers.

    Therefore: . P(n\text{ is even}) \:=\:\dfrac{750}{1470} \;=\;\dfrac{25}{49}




    \text{(b) }n\text{ is odd.}

    P(n\text{ is odd}) \;=\;1 - \dfrac{25}{49} \;=\;\dfrac{24}{49}




    \text{(c) }n\text{ is less than 4000.}

    The first digit must be 1, 2 or 3: .3 choices.
    The other three digits may be placed in 7\cdot6\cdot5 \:=\:210 ways.
    . . There are: . 3\cdot210 \:=\:630 numbers less than 4000.

    Therefore: . P(n < 4000) \;=\;\dfrac{630}{1470} \;=\;\dfrac{3}{7}




    \text{(d) }n\text{ is less than 4000, given that it is greater than 3000}

    We want: . P(n < 4000\,|\,n > 3000)

    \text{Bayes' Theorem: }\;P(n < 4000\,|\,n > 3000) \;=\;\dfrac{P([n < 4000]\, \wedge\, [n > 3000])}{P(n > 3000)} .[1]


    The numerator is: . P(3000 < n < 4000)

    The first number must be 3.
    The other three digits can be placed in 7\cdot6\cdot5 \:=\:210 ways.
    There are 210 numbers between 3000 and 4000.
    . . P(3000 < n < 4000) \:=\:\dfrac{210}{1470} \:=\:\dfrac{1}{7}


    The denominator is: . P(n > 3000)

    The first digit is 3, 4, 5, 6, or 7: .5 choices.
    The other three digits can be placed in 7\cdot6\cdot5 \:=\:210 ways.
    There are 5\cdot210 \:=\:1050 numbers greater than 3000.
    . . P(n > 3000) \:=\:\dfrac{1040}{1470} \;=\;\dfrac{5}{7}


    Substitute into [1]:

    . . P(n < 4000\,|\,n > 3000) \;=\;\dfrac{\frac{1}{7}}{\frac{5}{7}} \;=\;\dfrac{1}{5}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    This is why I HATE probability with a passion :P

    Thanks Soroban.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with probability problems!...
    Posted in the Statistics Forum
    Replies: 2
    Last Post: September 23rd 2010, 08:55 AM
  2. Probability problems
    Posted in the Statistics Forum
    Replies: 10
    Last Post: March 2nd 2010, 12:13 PM
  3. Need help: Probability Problems
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 20th 2009, 05:06 PM
  4. Problems about probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 24th 2009, 11:58 PM
  5. Probability Problems
    Posted in the Statistics Forum
    Replies: 2
    Last Post: March 8th 2009, 05:15 PM

Search Tags


/mathhelpforum @mathhelpforum