Download Statistics 1040 Review Exercises

Statistics 1040
Review Exercises - Chapter 15
Binomial Formula
Problem 1. Roll a die 6 times. Probability of exactly one ace = ??
On any one throw, P (Ace) = 1/6 and P (Non-Ace) = 5/6.
We can get EXACTLY one ace if we get:
a. ace on first throw AND non-ace on all other throws = P (ANNNNN) =
(1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = 3125 / 46,656 = 0.0670
OR
b. ace on second throw AND non-ace on all other throws = P(NANNNN) =
(5/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = 3125 / 46,656 = 0.0670
OR
c. ace on third throw AND non-ace on all other throws = P (NNANNN) = 0.0670
OR
c. ace on fourth throw AND non-ace on all other throws = P (NNNANN) = 0.0670
OR
c. ace on fifth throw AND non-ace on all other throws = P (NNNNAN) = 0.0670
OR
c. ace on sixth throw AND non-ace on all other throws = P (NNNNNA) = 0.0670
More simply, we can calculate the odds of one possible way of getting one ace and 5 non-aces,
and multiply by the number of possible combinations:
6 * P (ANNNNN) = 0.4019 or a bit over 40 percent.
The 6 is easily enough found, since the A can come in one of 6 positions. Note that the binomial formula will
do this for you, although it may be overkill for this problem:
Remember that my notation is C (n, k) = n ! / (k ! * (n - k) !)
C (6, 1) = 6 ! / (1 ! * 5 !) = 6 * 5 * 4 * 3 * 2 * 1 / 1 * (5 * 4* 3 * 2 * 1) = 6
Problem 2. Roll a die 10 times. What is the chance that it NEVER lands showing 6 ?
Note that no "OR" is involved -- it must land as a non-six every time, with chance 5/6 on any one
of the 10 independent tosses, so we can use the multiplication rule:
(5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6)
= (pow 5/6 10) = 9,765,625 / 60,466,176 = 0.1615
The chance of AT LEAST one 6 will be given by the complement rule as 1 - (pow 5/6 10) = 0.8385
Not asked in this problem, but worth thinking about:
What is the chance of exactly 3 sixes in the 10 rolls?
Note that approaching this with a complete listing of the possibilities, as in problem 1, would
take a long, long time. Fortunately, the answer is given by the binomial formula:
C (10, 3) * (pow 1/6 3) (pow 5/6 7)
If you extend this to calculating the probability of exactly 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, OR 10 sixes, you will
have constructed a binomial distribution.
Problem 3. Chance of more girls than boys in a family of 4 children.
Assume equal chance on any one draw, and independence of draws.
We are looking for the chance of 3 OR 4 girls.
Prob (exactly 4 girls) = (pow 1/2 4) = 1/16
Prob (exactly 3 girls and one boy) = C (4, 3) * (pow 1/2 3) * (pow 1/2 1) = 4/16
So the chance of 3 OR 4 girls will be 5 / 16
Note that the chance of 2 boys and 2 girls is = C (4, 2) * (pow 1/2 2) * (pow 1/2 2) = 6/16 = 0.3750
It is the most likely single outcome -- but its chance is less than 50 percent
Note also that P (more girls) + P (2 G and 2 B) + P (more boys) = 5/16 + 6/16 + 5/16 = 1.00 or 100 percent
Chapter 15 Review Exercises -- page 2
Problem 4. Box with 8 red marbles and 3 green ones; 6 draws without replacement.
What is the chance of exactly 3 green marbles?
If the problem had been draw 6 WITH replacement, the odds would not change with each draw and
the formula given would be correct:
C ( 6, 3) (pow 8/11 3) (pow 3/11 3) = 20 * (512/1331) * (27/1331) = 0.1561
But WITHOUT replacement, the draws are NOT INDEPENDENT. The binomial coefficient will
however still give us the possible combinations, and the odds of drawing 3 do not depend on the order of draws
P (G G G R R R) = P (G R G R G R) = P (R R R G G G) and so on.
Note that
P (G G G R R R) = (3 / 11) * (2 / 10) * (1 / 9) * (8 / 8) * (7 / 7) * ( 6 / 6) = 2016 / 332,640
and
P (G R G R G R ) = (3 / 11) * (8 / 10) * (2 / 9) * (7 / 8) * (1 / 7) * (6 / 6) = 2016 / 332,640
So the correct answer will be:
C ( 6, 3) * 2016 / 332,640 = 20 * 0.0061 = 0.1212 or 12.12 percent.
Problem 5. Committees from a club with 8 members.
Selecting all possible committees with 2 members is equivalent to asking for the number of
combinations of 8 people taken two at a time -- the answer is given by the binomial coefficient:
C ( 8, 2) = 8 ! / ( 2 ! * 6 !) = (8 * 7 * 6 * 5 * 3 * 2 * 1) / [ (2 * 1) * (6 * 5 * 4 * 3 * 2 * 1)]
= (8 * 7) / ( 2 * 1)
= 56 / 2 = 28
C (8, 5) = 8 ! / (5 ! * 3 !)
=
(8 * 7 * 6) / (3 * 2) = 56
It is tempting to think that with fewer people, there must be more choices. This is not necessarily true.
Problem 6. Committees from a club with 8 members.
Will the number of possible combinations get bigger if the size of the committee continues to increase?
C (8,2) = 8! / (2 ! * 6 !) = 28 from the last problem, but
C (8, 6) = 8! / (6! * 2!) = 28 as well.
What size committee gives you the MAXIMUM number of combinations?
Problem 7. Box with one red and 9 green marbles; 5 draws WITH replacement.
The chance of exactly two red marbles being drawn is as given:
C (5, 2) * (pow 1/10 2) * (pow 9/10 3) = 20 * (1/100) * (729 / 1000) = 10 * 729/100,000
The two last terms tell you the chance of the outcome red AND red AND green AND green AND green:
P (R R G G G) = 729 / 100,000
The combinations term does the job of the OR: you also get two reds if the outcome is
G R R G G or G G R R G or G G G R R (4 arrangements so far, with the two R's together)
or R G R G G or G R G R G or G G R G R (3 more with a green between the two R's)
or R G G R G or G R G G R (2 more with two green between the two R's)
or R G G G R (1 final arrangement with three green between the two R's)
C (5, 2) = 5 ! / (2 ! * 3 !) = (5 * 4) / (2 * 1) = 10 tells you the number of such arrangements,
and relieves you from the worry that you might have missed one or two possibilities.
It plays the role of the "OR" -- multiplying by 10 is the same as adding 10 probabilities.
Chapter 15 Review Exercises -- page 3
Problem 8. Coin tossed 10 times. Find chance of exactly 2 heads among first 5 tosses and exactly 4 heads
among the last 5 tosses.
P (2 heads in 5 tosses) = C (5, 2) * (pow 1/2 2) * (pow 1/2 3) = 5! / (2! * 3!) * 1/4 * 1/8
= 10 / 32
P (4 heads in 5 tosses) = C (5, 4) * (pow 1/2 4) * (pow 1/2 1) = 5! / (4! * 1!) * 1/16 * 1/2
= 5 / 32
The chance that both happen is given by the simple multiplication rule, since the outcomes are
independent:
P (2 heads in first 5 tosses AND 4 heads in next 5) = (10/32) * (5/32) = 50 / 1024 = 0.0488
Problem 9. Drawing cards, and thinking about EXCLUSIVE, INDEPENDENT and which rule to apply.
a. Chance that the top card is K of spades AND bottom card is Q of spades
Multiplication rule for DEPENDENT events (drawing the K on top means the Q is NOT on top) =
P(KofS on top) * P (Qof S on bottom | KofS on top) = (1/52) * (1/51)
b. Chance that the top card is the K of spades AND the bottom card is the K of spades
You could realize that both cannot possibly happen with a fair deck, or you could formalize this
as another example of the multiplication rule for DEPENDENT events:
P (KofS on top) * P (KofS on bottom | KofS on top) = (1/52) * (0/52) = zero
Since the two events are EXCLUSIVE, they are NOT independent; the chance of two exclusive
events happening is zero.
c. Chance that the top card is the K of spades OR the bottom card is the K of spades.
The addition rule for exclusive events applies:
P (KofS on top) + P (KofS on bottom) = 1/52 + 1/52
d. Chance that the top card is the K of spades OR the bottom card is the Q of spades.
These two events are NOT exclusive, so we must use the full addition rule:
P (KofS on top) + P (QofS on bottom) - P (KofS on top AND QofS on bottom)
= 1/52 + 1/52 - (1/52 * 1/51)
This can be put in the form of an option on the textbook list, after some manipulation:
= 51 / 51*52 + 52 / 51*52 - 1 / 51* 52
= 102 / 51*52 = (51 + 51) / 51*52 = 1/52 + 1/52 which is option (iv).
Only a real meanie would ask you to do this on an exam.
Problem 10. Drawing with or without replacement makes a difference !
Box has 3 red tickets and 2 green ones.
You win $ 1 if in five draws 3 tickets are red and 2 are green.
With replacement, the chance of winning is:
C (5, 3) * (pow 3/5 3) * (pow 2/5 2) = [5! / (3! * 2!)] * (27/125) * 4/25 = 1080 / 3125
= 0.3456 or 34.56 percent.
Without replacement, all balls in the box are drawn, and your chances of winning are 100 percent !
Problem 10.5 (added) The Lady Tasting Tea
Sir Ronald Fisher met a woman who claimed she could tell whether milk or tea had been poured into the cup
first. He tested her ability to do so on four cups, and she got three out of the four right. Does this prove she had
the ability she claimed? Probability of 4 right by guessing = (pow 1/2 4) = 1/16 makes it unlikely that you
would get all right, but probability of 3 right = C (4,3) * (pow 1/2 3) * (pow 1/2 1) = 4/16 = 0.25 doesn't very
strongly reject the hypothesis she was guessing. See David Salsburg, The Lady Tasting Tea: How Statistics
Revolutionized Science in the Twentieth Century (NY, Freeman, 2001).
Chapter 15 Review Exercises -- page 4
Problem 11 -- Smoking and Health
[Complex problem, which anticipates most of the rest of the semester]
Sir Ronald Fisher questioned the early research on smoking and health, noting that an alternative hypothesis was
that there could be a common genetic factor which predisposed to both smoking and lung cancer or heart
disease. Research on twins was done to exclude genetic factors, so any difference between death rates of twins
would have to be due to either random chance or to harm done by smoking. But as in the case of the lady tasting
tea, Fisher could ask whether the evidence was strong enough to reject the hypothesis of pure chance.
a. Consider the fact that the smoking twin died first in 17 of the 22 cases observed. What is the probability of
this happening by chance if both twins really had an equal chance of dying first?
First, consider one possible arrangement of the deaths (the first 17 pairs have the smoker die first, the last 5 the
non-smoker):
P (S S S S S S S S S S S S S S S S S N N N N N) = (pow 1/2 17) * (pow 1/ 2 5) = 1/ 4,194,304
= 0.0000002384 or .00002384 percent.
Since there are C (22, 17) = 22 ! / (17 ! 5!) = (22 * 21 * 20 * 19 * 18) / (5 * 4 * 3 * 2 * 1)
= 26,334 possible combinations of the 17 smokers and 5 non-smokers,
the chance of 17 smokers dying first by chance is
26,334 / 4,194,304 = 0.0062785149 or 0.6278 percent
There is just a bit over half a percent probability of this happening by chance.
The computer will save a lot of work with the DBINOM (binomial density function) which gives the probability
of a given outcome happening k times in n trials, if the probability of that outcome is p:
(dbinom k n p) = (dbinom 17 22 1/2) = 0.0062785149
We could (and should by the strict rules of hypothesis testing) ask the question whether it is possible for that
result or an even more unfavorable result (18,19,20,21 or all 22 smokers dying first) to happen by chance.
(dbinom 18 22 1/2) = 0.0017440319
(dbinom 19 22 1/2) = 0.0003671646
(dbinom 20 22 1/2) = 0.0000550747
(dbinom 21 22 1/2) = 0.0000052452
(dbinom 22 22 1/2) = 0.0000002384
But adding all these up brings the chance only to 0.0084502697, or 0.8450 percent -- still less than one percent.
The command (binomial 22 .5) will show you the entire distribution; you can place a normal curve over it by
pressing the "n" key.
b. For the heart deaths, all 9 smokers died first. The probability of this happening by chance is
(pow 1/2 9) = 1 / 512 = 0.00195 or about 0.2 percent -- two-tenths of one percent.
This is strong evidence against the heart deaths being due to chance -- in fact, even stronger than was the
case with the overall death rate in part a.
c. For the lung cancer deaths, we only have 2 deaths -- and the probability of that happening by chance is
the same as the chance of two tosses of a fair coin coming up heads = 1/4. There is not enough evidence
from this to reject the hypothesis that lung cancer deaths were due to chance rather than smoking.