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BINOMIAL DISTRIBUTION

VIEWS: 7 PAGES: 14

									                                  DISTRIBUTIONS

1. Toss a coin 3 times. Find the probability of getting 2 heads.

2. Draw 6 cards from a deck of 52 cards without replacement. What is the
   probability of getting two hearts?

3. A printer in a computer lab works 60% of the time. Find probability that the
   printer is working 2 out of the 10 visits.

4. Suppose you invest a fixed sum of money in each of five business ventures. Each
   has a 70% chance of success. Find the probability that exactly 2 of the business
   ventures will be successful.

5. A large population of accounts is known to have 10% of accounts in error.
   Calculate the probability that when accounts are chosen at random from this
   population, the first account chosen will be in error.

6. A rail company knows that on average they will have 3 people ill each day. In
   order to make plans for replacements they have asked you to calculate the
   following probabilities.
   In a single day
       (i)     No employees ill
       (ii)    Two or less employees ill
       (iii) Four or more employees ill

7. Suppose the probability that a married couple will have child with blonde hair is
   0.25. If there are four children in the family, what is the probability that exactly
   half of them have blonde hair.

8. On the bus route that a student takes to college, there are 5 traffic lights between
   the stop the student boards the bus and the stop the student embarks. Assume that
   the probability that a light will be red is 0.25. Assuming independence, what is
   the probability that at least three traffic lights will show a red light before the
   student gets off the bus?

9. A secretary receives application forms for a current vacancy and has to check
   them for completeness before further processing is carried out on them. Suppose
   in a batch of 20, 7 are incomplete and the others are complete. If the secretary
   randomly selects 5 of these forms for checking, what is the probability that:
   i) none of the incomplete forms are selected?
   ii) at least 3 of these selections are incomplete?

10. Jack applied for 8 jobs as an accountant last week. The probability
    of being called for an interview is 0.25. Assuming independence,
      (i)    what is the probability that Jack will be called for at most
             2 interviews.
      (ii)   on average, of the 8 jobs Jack applied for, how many
             interviews should Jack expect to be called for?
11. On average 2 people are killed on a particularly dangerous stretch of road each
    year.
      a) What is the probability that in a given year that
          (i) No one is killed
          (ii) 2 or less are killed
          (iii)Four or more are killed
      b) What is the probability that over a 3 year period less than 3 people are
          killed

12. A company tenders for a new contract each week. In the past they have won
    40% of the contracts they have tendered for. In a 6-week period what is the
    probability that they will
      (i)    Win all six contracts
      (ii)   Win at least 2 contracts
      (iii) Win less than 3 contracts

13. The number of blemishes that occur on a roll of carpet during manufacture is
    Poisson distributed, with mean 0.4 blemishes per roll. What percentage of rolls
    are classified
      (i)     Perfect (having no blemishes)
      (ii)    Seconds (having one or two blemishes)
      (iii) Scrap (all the rest)
1.    Bin
2.    Hypergeometric
3.    Bin
4.    Bin
5.    Geometric
6.    Poisson
7.    Bin
8.    Bin
9.    Hypergeometric
10.   Bin
11.   Poisson
12.   Bin
13.   Poisson
                                  DISTRIBUTIONS
Binomial
1. Toss a coin 3 times. Find the probability of getting 2 heads.

2. Suppose you invest a fixed sum of money in each of five business ventures. Each
   has a 70% chance of success. Find the probability that exactly 2 of the business
   ventures will be successful.

3. A printer in a computer lab works 60% of the time. Find probability that the
   printer is working 2 out of the 10 visits.

4. A stock will show an increase or a decrease in its closing price on a daily basis.
   With this particular stock, the probability of an increase is 0.65.
   What is the probability that there will be at least 1 increase in the stock price in
   the next ten days.

5. A company tenders for a new contract each week. In the past they have won 40%
   of the contracts they have tendered for. In a 6-week period what is the probability
   that they will
       (iv)    Win all six contracts
       (v)     Win at least 2 contracts
       (vi)    Win less than 3 contracts

6. Suppose the probability that a married couple will have child with blonde hair is
   0.25. If there are four children in the family, what is the probability that exactly
   half of them have blonde hair.

7. On the bus route that a student takes to college, there are 5 traffic lights between
   the stop the student boards the bus and the stop the student embarks. Assume that
   the probability that a light will be red is 0.25. Assuming independence, what is
   the probability that at least three traffic lights will show a red light before the
   student gets off the bus?

8. Jack applied for 8 jobs as an accountant last week. The probability
   of being called for an interview is 0.25. Assuming independence,
     (iii)  what is the probability that Jack will be called for at most
            2 interviews.
     (iv)   on average, of the 8 jobs Jack applied for, how many
            interviews should Jack expect to be called for?
Hypergeometric
8. Draw 6 cards from a deck of 52 cards without replacement. What is the
   probability of getting two hearts?

9. A secretary receives application forms for a current vacancy and has to check
   them for completeness before further processing is carried out on them. Suppose
   in a batch of 20, 7 are incomplete and the others are complete. If the secretary
   randomly selects 5 of these forms for checking, what is the probability that:
   ii) none of the incomplete forms are selected?
   ii) at least 3 of these selections are incomplete?

Geometric
10. A large population of accounts is known to have 10% of accounts in error.
    Calculate the probability that when accounts are chosen at random from this
    population, the first account chosen will be in error.

11. Auditions for a new musical will last on average 20 minutes. The director in
    charge of the audition knows from past experience that the probability that a
    person is suitable for the part is 0.8. Due to time constraints, the first person
    auditioned, that the director deems suitable, is selected. If the talent for any one
    person is independent of another, what is the probability that the director will have
    found the person for the part in exactly one hour after starting the auditions?

Poisson
12. A rail company knows that on average they will have 3 people ill each day. In
    order to make plans for replacements they have asked you to calculate the
    following probabilities.
        In a single day
        (iv)    No employees ill
        (v)     Two or less employees ill
        (vi)    Four or more employees ill

13. On average 2 people are killed on a particularly dangerous stretch of road each
    year.
      a) What is the probability that in a given year that
          (iv) No one is killed
          (v) 2 or less are killed
          (vi) Four or more are killed
      b) What is the probability that over a 3 year period less than 3 people are
          killed

14. The number of blemishes that occur on a roll of carpet during manufacture is
    Poisson distributed, with mean 0.4 blemishes per roll. What percentage of rolls
    are classified
      (iv)    Perfect (having no blemishes)
      (v)     Seconds (having one or two blemishes)
      (vi)    Scrap (all the rest)
     2006 Exam




2005 Repeat Exam

Q4
2004 Exam
Question 3




2004 Exam
Question 4




2004 Repeat Exam
Question 4
  Solution to question 5

  M = 13 number of hearts

  L = 39 number of non-hearts

  N = 52 total


               13  39 
                 
                2  4 
P (2 hearts)     =0.315
                   52 
                   
                  6 
                   
BINOMIAL DISTRIBUTION
Q3
     If the probability that any one person requires medical attention in one
     particular year is 0.25, what is the probability that:
     (i)     four people live through the year without medical attention?
     (ii)    just one of the four needs medical attention?
     (iii) at least two of the four require medical attention

Q4
       A company tenders for a new contract each week. In the past they have won
       40% of the contracts they have tendered for. In a 6-week period what is the
       probability that they will
       (vii) Win all six contracts
       (viii) Win at least 2 contracts
       (ix)   Win less than 3 contracts

Q5
       Suppose the probability that a married couple will have child with blonde hair
       is 0.25. If there are four children in the family, what is the probability that
       exactly half of them have blonde hair.



POISSON DISTRIBUTION
Q6
     A rail company knows that on average they will have 3 people ill each day. In
     order to make plans for replacements they have asked you to calculate the
     following probabilities.

       In a single day
       (vii) No employees ill
       (viii) Two or less employees ill
       (ix)    Four or more employees ill

Q7
       On average 2 people are killed on a particularly dangerous stretch of road each
       year.
       a) What is the probability that in a given year that
       (vii) No one is killed
       (viii) 2 or less are killed
       (ix)   Four or more are killed
       b) What is the probability that over a 3 year period
       (i)    Less than 3 people are killed

Q8
       The number of blemishes that occur on a roll of carpet during manufacture is
       Poisson distributed, with mean 0.4 blemishes per roll. What percentage of rolls
       are classified
(vii) Perfect (having no blemishes)
(viii) Seconds (having one or two blemishes)
(ix)   Scrap (all the rest)
BINOMIAL DISTRIBUTION
Q3
If the probability that any one person requires medical attention in one particular year
is 0.25, what is the probability that:
        (i)     four people live through the year without medical attention?
        (iv)    just one of the four needs medical attention?
        (v)     at least two of the four require medical attention

We define a person requiring medical attention as a “success”, and the probability of a
success equals 0.25.

Using the Binomial Distribution
                    n!      X        n- X
        P(X) =             p (1 - p )
               X! (n - X)!
with
n=4 people
p=0.25
1-p =0.75

(i) No-one needs medical attention P(X=0)
                       4!                            4!
                               (0.25) ( 0.75 )           (1)(.32)  .32
                                     0        4- 0
        P( 0 ) =
                 0! ( 4 - 0 )!                     1.(4!)

(ii) P(X =1)

                         4!                           4.3.2.1
                                 (0.25) ( 0.75 )              (0.25)(.42)  .42
                                       1        4- 1
        P( 1 ) =
                   1! ( 4 - 1 )!                     1.(3.2.1)

(iii) At least two need medical attention P(X2)
         P(X2) =P(2) + P(3) + P(4) = 1 –( P(0) + P(1))
                                     = 1 –(.32+.42) = .26

Q4
Winning a contract can be defined a success. The probability of winning a contract (p)
= .4. There are six contracts (n=6).

Using Binomial
                        n!      X        n- X
        P(X) =                 p (1 - p )
                   X! (n - X)!

                        6!          0      6 0
        P( 0 ) =                 .4 ( .6 ) = .047
                  0! ( 6 - 0 )!
                       6!         1     6 1
        P( 1 ) =               .4 ( .6 ) = .187
                 1! ( 6 - 1 )!
                        6!          2      6 2
        P( 2 ) =                 .4 ( .6 ) = .311
                  2! ( 6 - 2 )!
                       6!         3     6 3
        P( 3 ) =               .4 ( .6 ) = .276
                 3! ( 6 - 3 )!
                       6!          4      6 4
        P( 4 ) =                .4 ( .6 ) = .138
                 4! ( 6 - 4 )!
                       6!         5      6 5
        P( 5 ) =               .4 ( .6 ) = .037
                 5! ( 6 - 5 )!
                       6!          6     6 6
        P( 6 ) =                .4 ( .6 ) = .004
                 6! ( 6 - 6 )!

(i) Win all six contracts P(X= 6)
       P(X=6) = .004

(i)    Win at least 2 contracts
       P(X2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)
              =(.311 + .276 + .138 + .037 + .004) = .766
       OR
       P(X2) = 1 – P(X< 2) 1 - P(0) + P(1) = 1 – (.047 + .187) = .766

(iii) Win less than 3 contracts
        P(X< 3) = P(X=0) + P(X=1) + P(X=2) = .05 +.19 + .31 = .55


Q5
Probability of blond hair = .25. There are 4 kids (n=4). What is the probability that
half of them (X=2) have blond hair.

                        n!      X        n- X
        P(X) =                 p (1 - p )
                   X! (n - X)!

                      4!
                             .25 ( .75 )  6(.0625)(.625)  .21
                                2       2
        P( 2 ) =
                   2! ( 2 )!



POISSON DISTRIBUTION
Q6

       (i)         No employees ill, X = 0, =3
                          e 3 30 (.05)(1)
                   P(0)                   = .05
                             0!       1

       ii)         Two or less ill
                   P(X2) = P(0) + P(1) + P(2)

                            e 3 (3)1 (.05).3
                   P(1)                     = .15
                                1!       1
                        e 3 (3) 2 (.05)(9)
               P(2)                        .22
                            2!        2

               P(X2) = P(0) + P(1) + P(2)= .05 +.15 + .22 =.42

       (iii)   Probability of four or more employees ill
               P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3))

                        e 3 (3) 3 (.05)(27)
               P(3)                         =.22
                            3!         6

               P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3))

               = 1 – (.05 +.15 +.22 +.22) = 1 - .64 =.36


Q7
a) On average 2 people are killed (=2)

       (i)     No-one killed, X = 0, =2

                        e 2 2 0 (.135)(1)
               P(0)                      = .135
                           0!        1

       ii)     Two or less killed

               P(X2) = P(0) + P(1) + P(2)

                        e 2 (2)1 (.135).2
               P(1)                      = .27
                            1!       1

                        e 2 (2) 2 (.135)(4)
               P(2)                         .27.
                            2!         2

               P(X2) = P(0) + P(1) + P(2) = .135 +.27 + .27 =.675

       (iii)   Probability of four or more killed
               P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3))

                        e 2 (2) 3 (.135)(8)
               P(3)                         =.18
                            3!         6

               P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3))

               = 1 – (.135 +.27 +.27 +.18) = 1 - .855 =.145
b) In a 1-year period 2 people are killed on average this implies in a 3 year period 6
people are killed on average. (=6)

(i) Prob (X < 3)

                                 e 6 (6) 0 (.002)(1)
                        P(0)                         .002
                                     0!         1

                                 e 6 (6)1 (.002)(6)
                        P(1)                       = .012
                                     1!        1

                               e 6 (6) 2 (.002)(36)
                        P(2)                        .036
                                   2!         2

        Prob (X < 3) = P(0) + P(1) + P(2) = .002 +.012 + .036 = .05


Q8
        Mean number of blemishes = 0.4 (=0.4)

(i)     Perfect (having no blemishes) P(X = 0)

                 e 0.4 (0.4) 0 (.6703)(1)
        P(0)                              .6703 (67.03% will be perfect)
                        0!          1

(ii)    Seconds (having one or two blemishes) P(X = 1 or X = 2)

               e 0.4 (0.4)1 (.6703)(0.4)
        P(1)                              . 2561
                      1!           1
               e 0.4 (0.4) 2 (.6703)(.16)
        P(2)                              .0536
                       2!           2

P(X = 1 or X = 2) = P(1) + P(2) = .2561 + . 0536 = .3097 (30.97% will be seconds)

(iii)   Scrap (all the rest) P(X > 2)

P(X > 2) = 1 – (P(0) + P(1) + P(2)) = 1 – (.6703 + .2561 + .0536) = 1 - .98 = .02
(2% of all carpets will be scrap)

								
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