Results 1 to 8 of 8

Math Help - Binomial Distribution?

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    52

    Binomial Distribution?

    Hi, should I use binomial distribution to solve the following question? (a) to be specific. I managed to solve (b)

    Research shows that 40% of Singaporeans prefer tea, 35% prefer coffee and the rest prefer milk. 9 persons were randomly chosen. Find the probability that

    (a) the same number of people prefer each type of drink. Ans:0.07203

    (b) more people prefer tea to the other drinks. Ans:0.2666

    T~B(9,0.4)
    C~B(9,0.35)
    M~B(9,0.25)

    P(T=3) x P(C=3) x P(M=3)
    = {9C3(0.4)^3 x (0.6)^6} x {9C3(0.35)^3 x (0.65)^6} x {9C3(0.25)^3 x (0.75)^6}
    = 0.01591

    But the answer given is 0.07203. Spot any mistakes?
    Thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by cyt91 View Post
    Hi, should I use binomial distribution to solve the following question? (a) to be specific. I managed to solve (b)

    Research shows that 40% of Singaporeans prefer tea, 35% prefer coffee and the rest prefer milk. 9 persons were randomly chosen. Find the probability that

    (a) the same number of people prefer each type of drink. Ans:0.07203

    (b) more people prefer tea to the other drinks. Ans:0.2666

    T~B(9,0.4)
    C~B(9,0.35)
    M~B(9,0.25)

    P(T=3) x P(C=3) x P(M=3)
    = {9C3(0.4)^3 x (0.6)^6} x {9C3(0.35)^3 x (0.65)^6} x {9C3(0.25)^3 x (0.75)^6}
    = 0.01591

    But the answer given is 0.07203. Spot any mistakes?
    Thanks for your help.
    Multinomial distribution - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    652
    Thanks
    128
    You can get there using the Binomial Distribution.

    The probability that precisely 3 people prefer tea is 9C3(0.4)^{3}(0.6)^{6}.

    The probabilty that of the remaining 6, precisely 3 prefer coffee is 6C3(35/60)^{3}(25/60)^{3}.

    Since you need both of these to happen you simply multiply them. You don't have to worry about the 'milkies' they will just be the remaining 3.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Thanks for your reply. Why can't I reach the same answer with my (wrong) method?
    What is the problem with my approach?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    Posts
    652
    Thanks
    128
    Basically, you are counting the same people twice or three times.
    Your line P(T=3)*P(C=3)*P(M=3) should be read as, the probability that exactly three people drink tea AND THEN (of the remainder), exactly three people drink coffee AND THEN (of the remainder) exactly three people drink milk.
    The final probability P(M=3) will be equal to 1 because the tea and coffee drinkers have been removed, the three that are left are bound to prefer milk. The probability for the coffee drinkers is of a three from six situation.
    You should be able to arrive at the same result by carrying out the calculation in a different order. That is, working out say (using your notation) P(M=3)*P(C=3)*P(T=3) or P(C=3)*P(M=3)*P(T=3) etc (six possible different orders) should always give you the same result, and in each case the third of the probabilities will be 1.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Thanks a lot. You've been very helpful.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, cyt91!

    Should I use binomial distribution to solve the following question?

    With three events, we must use a multinomial distribution.



    Research shows that 40% prefer tea, 35% prefer coffee and the rest prefer milk.
    Nine persons were randomly chosen. .Find the probability that

    (a) the same number of people prefer each type of drink. .Ans: 0.07203

    \displaystyle P(\text{3 of each}) \;=\;{9\choose3,3,3}(0.40)^3(0.35)^3(0.25) \;=\;0.07203



    Your answer should have been derived something like this . . .

    We want three of the nine people to drink Tea: . \displaystyle {9\choose3}(0.4)^3

    Of the remaining six people,
    . . we want three Coffee and three Milk: . \displaystyle {6\choose3}(0.35)^3(0.25)^3

    The answer is: . \displaystyle {9\choose3}(0.6)^3{6\choose3}(0.35)^3(0.25)^3 \;=\;0.07203




    (b) more people prefer tea to the other drinks. . Ans:0.2666

    P(\text{Tea}) \:=\:0.4,\;\;P(\text{Other}) \:=\:0.6


    \begin{array}{ccccc}<br />
P(\text{9 Tea, 0 Other}) &=& {9\choose9}(0.4)^9(0.6)^0 &=& 0.000\,262\,144 \\ \\ [-3mm]<br />
P(\text{8 Tea, 1 Other}) &=& {9\choose8}(0.4)^8(0.6)^1 &=& 0.003\,538\,944 \\ \\ [-3mm]<br />
P(\text{7 Tea, 2 Others}) &=& {9\choose7}(0.4)^7(0.6)^2 &=& 0.021\,233\,664 \\ \\ [-3mm]<br />
P(\text{6 Tea, 3 Others}) &=& {9\choose6}(0.4)^6(0.6)^3 &=& 0.074\,317\,824 \\ \\ [-3mm]<br />
P(\text{5 Tea, 4 Others}) &=& {9\choose5}(0.4)^5(0.6)^4 &=& 0.167\,215\,104 \\ \\[-4mm] \hline \\[-4mm]<br />
&& \text{Total:} && 0.266\,567\,680\end{array}


    Therefore: . P(\text{more Tea}) \:\approx\:0.2666
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Thanks a lot. You guys have been extremely helpful. Keep up the good work!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Binomial Distribution
    Posted in the Statistics Forum
    Replies: 6
    Last Post: November 18th 2011, 12:48 PM
  2. Replies: 3
    Last Post: March 21st 2010, 05:25 PM
  3. Replies: 1
    Last Post: November 12th 2009, 12:38 AM
  4. Replies: 1
    Last Post: March 11th 2009, 11:09 PM
  5. Cumulative distribution function of binomial distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 31st 2008, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum