Fall 2019 MTH 115 Take-Home Exam: Statistics and Probability Problems

This document presents the solutions to a take-home exam for a statistics course (MTH 115), focusing on probability and data analysis. The exam covers various concepts including basic probability calculations, conditional probability, and dependent/independent events. Problems involve analyzing data from the 2018 Winter Olympics medal table, calculating probabilities of drawing specific medals, and determining the likelihood of certain events. The solutions also address binomial distributions, permutations, and the application of statistical concepts to real-world scenarios. The student provides detailed reasoning and calculations for each problem, demonstrating a thorough understanding of the material. The exam also delves into questions about unusual occurrences in binomial distribution, permutations, and combinations. The document provides a comprehensive analysis of each question, demonstrating a strong grasp of statistical concepts.

STATISTICS
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Question 1
Requisite probability = (6/142) = 0.042
Question 2
Probability of drawing first silver = (46/142)
Probability of drawing second silver (without replacement) = (45/141)
Requisite probability = (46/142) * (45/141) = 0.1034
Question 3
Favourable cases = 39 (Total Norweign Medals) + 56 (All Gold) – 14 (Gold from Norway
since they have been double counted) = 81
Requisite probability = (81/142) = 0.57
Question 4
Requisite probability for first German medal = (31/142)
Probability for second German medal (without replacement) = (30/141)
Probability for third German medal (without replacement) = (29/140)
Requisite probability = (31/142)*(30/141)*(29/140) = 0.0096
Question 5
Probability of bronze medal in first draw = (40/142)
Probability of bronze medal in second draw (without replacement) = (39/141)
Requisite probability = (40/142)*(39/141) = 0.0779
Question 6
The given events are not independent but are rather dependent events. This is because the
pushing out of the opponents stone from the button is dependent on the stone landing in the
button. If the stone would land outside the button, then it cannot push the opponents stone
already present in the button.
Question 7
Yes, the two events do represent disjoint sets. This is because the two teams considered are
representing different countries and hence there would not be any member who is
representing two countries and that too in different sports.
Question 8

If the first pole is red, then probability = (5/10)
Then second pole has to be blue which has probability (without replacement) = (5/9)
Third pole has to be red which has probability (without replacement) = (4/8)
Fourth pole has to be blue which has probability (without replacement) = (4/7)
Fifth pole has to be red which has probability (without replacement) = (3/6)
Sixth pole has to be blue which has probability (without replacement) = (3/5)
Seventh pole has to be red which has probability (without replacement) = (2/4)
Eighth pole has to be blue which has probability (without replacement) = (2/3)
Ninth pole has to be red which has probability (without replacement) = (1/2)
Tenth pole has to be blue which has probability = 1
Requisite probability = (5/10)*(5/9)*(4/8)*(4/7)*(3/6)*(3/5)*(2/4)*(2/3)*(1/2) = 0.00397
Since, the starting can be with red or blue, hence total probability = 0.00397 * 2 = 0.0079
Question 9
Requisite probability = (6!*1!*1!*3!/10!) = 0.00119
Question 10
Requisite probability from table = P(9) = 0.000181
Question 11
Requisite probability = P(0) + P(1) + P(2) + P(3) = 0.04344 + 0.155707 + 0.255804 +
0.254696 = 0.709647
Question 12
Method 1
Requisite probability = 1 –P(0) = 1-0.04344 = 0.95656
Method 2
Requisite probability = P(1)+ P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) +
P(11)+ P(12) = 0.95656
Question 13
Requisite probability (i.e. P>10) = P(11) + P(12) = 0.000001 + 0.000000 = 0.000001
Question 14

Success occurs when the selected American skier is a cross country skier. Failure occurs
when the selected American skier is not a cross country skier.
P = 0.23
Q = (1-p) = 1-0.23 = 0.77
N =12 (i.e. number of trials)
X = 6 (i.e. number of successes)
Requisite probability = 12C6 *(0.23)6*(0.77)12-6 = (12*11*10*9*8*7) /(1*2*3*4*5*6) *
0.000148 *0.208422 = 0.028509
Question 15
Mean = np = 12*0.23 = 2.76
Standard deviation = (npq)0.5= (12*0.23*0.77)0.5 = 1.458
Let the requisite number of cross country skiers amongst the 12 skiers be X.
For X to be unusually large, P(x>X) =0.05
Hence, the corresponding z value is 1.6448
1.6448 = ( X- 2.76)/(1.458/120.5)
Solving the above X = 3.45
Thus, at a minimum 4 skiers should be cross country for the incidence to be abnormal.
Question 16
Since for each of the numbers there are 40 options (0 to 39), hence the total passwords
possible are = 40*40*40 = 64,000
It is evident that for the above password, the order of digits is imperative. As a result, the
name combination lock is a misnomer since typically in combinations, the arrangement or
order does not matter. As a result, it would be more appropriate to label this as an
arrangement lock or permutation lock.