Biostatistics for the Health Sciences: Basic Probability - Exercises questions answers

**EXERCISES**

**5.1** By using a computer algorithm, an investigator can
assign members of twin pairs at random to an intervention condition in a
clinical trial. Assume that each twin pair consists of dizygotic twins (one
male and one female). The probability of assigning one member of the pair to
the intervention condition is 50%. Among the first four pairs, what is the
probability of assigning to the intervention condition: 1) zero females, 2) one
female, 3) two females, 4) three females, 4) four females?

**5.2** In this exercise, we would like you to toss four
coins at the same time into the air and record and observe the results obtained
for various numbers of coin tosses. Count the frequencies of the following
outcomes: 1) zero heads, 2) one head, 3) two heads, 4) three heads, 5) four
heads.

a. Toss the coins one time (and
compare to the results obtained in Exercise 5.1).

b. Toss the coins five times.

c. Toss the coins 15 times.

d. Toss the coins 30 times.

e. Toss the coins 60 times.

**5.3** In the science exhibit of a museum of natural
history, a coin-flipping machine tosses a silver dollar into the air and
tallies the outcome on a counting device. What are all of the respective
possible outcomes in any three consecutive coin tosses? In any three
consecutive coin tosses, what is the probability of: a) at least one head, *b*) not more than one head, c) at least
two heads, d) not more than two heads, e) exactly two heads, f) exactly three
heads.

**5.4** A certain laboratory animal used in preclinical
evaluations of experimental catheters gives birth to only one offspring at a
time. The probability of giving birth to a male or a female offspring is
equally likely. In three consecutive pregnancies of a single animal, what is
the probability of giving birth to: (a) two males and one female, (b) no
females, (c) two males first and then a fe-male, and (d) at least one female.
State how the four probabilities are different from one another. For the
foregoing scenario, note all of the possible birth outcomes in addition to (a)
through (d).

**5.5** What is the expected distribution—numbers and
proportions—of each of the six faces (i.e., 1 through 6) of a die when it is
rolled 1000 times?

**5.6** A pharmacist has filled a box with six different
kinds of antibiotic capsules. There are a total of 300 capsules, which are
distributed as follows: tetracy-cline (15), penicillin (30), minocycline (45),
Bactrim (60), streptomycin (70), and Zithromax (80). She asks her assistant to
mix the pills thoroughly and to withdraw a single capsule from the box. What is
the probability that the cap-sule selected is: a) either penicillin or
streptomycin, *b*) neither Zithromax
nor tetracycline, c) Bactrim, d) not penicillin, e) either minocycline,
Bactrim, or tetracycline?

**5.7** In an ablation procedure, the probability of acute
success (determined at com-pletion of the procedure) is 0.95 when an image
mapping system is used. Without the image mapping system, the probably of acute
success is only 0.80. Suppose that Patient A is given the treatment with the
mapping system and Patient *B* is given
the treatment without the mapping system. Determine the following
probabilities:

a. Both patients *A* and *B* had acute successes.

b. A had an acute success but *B* had an acute failure.

c. B had an acute success but A had an acute failure.

d. Both *A* and *B* had acute failures.

e. At least one of the patients had an acute success.

f. Describe two ways that the result in (e) can be calculated based on the
re-sults from (a), (b), (c), and (d).

**5.8** Repeat Exercise 5.4 but this time assume that the
probability of having a male offspring is 0.514 and the probability of having a
female offspring is 0.486. In this case, the elementary outcomes are not
equally likely. However, the trials are Bernoulli and the binomial distribution
applies. Use your knowledge of the binomial distribution to compute the
probabilities [(a) through (e) from Exercise 5.5].

**5.9** Refer to Formula 5.7, permutations of *r* objects taken from *n* objects. Compute the following
permutations:

a. *P*(8, 3)

b. *P*(7, 5)

c. *P*(4, 2)

d. *P*(6, 4)

e. *P*(5, 2)

**5.10 **Nine volunteers wish to participate in a clinical
trial to test a new medication** **for
depression. In how many ways can we select five of these individuals for
assignment to the intervention trial?

**5.11 **Use Formula 5.8, combinations of** ***r*** **objects taken out of** ***n*,
to determine the** **following
combinations:

a. *C*(7, 4)

b. *C*(6, 4)

c. *C*(6, 2)

d. *C*(5, 2)

e. What is the relationship between 5.11 (d) and 5.9 (e)?

f. What is the relationship between 5.11 (b) and 5.9 (d)?

**5.12 **In how many ways can four different colored marbles
be arranged in a row?

**5.13 **Provide definitions for each of these terms:

a. Elementary events

b. Mutually exclusive events

c. Equally likely events

d. Independent events

e. Random variable

**5.14 **Give a definition or description of the following:

a. *C*(4, 2)

b. *P*(5, 3)

c.The addition rule for mutually exclusive events

d. The multiplication rule for independent events

**5.15 **Based on the following table of hemoglobin levels
for miners, compute the** **probabilities
described below. Assume that the proportion in each category for this set of 90
miners is the true proportion for the population of miners.

a. Compute the probability that a miner selected at random from the
population has a hemoglobin level in the 12.0–17.9 range.

b. Compute the probability that a miner selected at random from the
population has a hemoglobin level in the 18.0–21.9 range.

c. Compute the probability that a miner selected at random from the
population has a hemoglobin level in the 22.0–27.9 range.

d. What is the probability that a miner selected at random will have a
hemoglobin count at or above 18.0?

e. What is the probability that a miner selected at random will have a
hemoglobin count at or below 21.9?

f. If two miners are selected at random from the “infinite population” of
min-ers with the distribution for the miners in the table, what is the
probability that one miner will fall in the lowest class and the other in the
highest (i.e., one has a hemoglobin count in the 12.0 to 17.9 range and the
other has a hemoglobin count in the 22.0 to 27.9 range)?

**5.16 **Consider the following 2 × 2 table that shows
incidence of myocardial infarc-tion (denoted MI) for women who had used oral
contraceptives and women who had never used oral contraceptives. The data in
the table are fictitious and are used just for illustrative purposes.

Assume that the proportions in
the table represent the “infinite population” of adult women. Let *A* = {woman used oral contraceptives} and
let *B* = {woman had an MI episode}

a. Find *P*(*A*), *P*(*B*), *P*(*A ^{c}*), and

b. What is *P*(*A **∩** B*)?

c. What is *P*(*A U B*)?

d. Are *A* and *B* mutually exclusive?

e. What are *P*(*A*|*B*)
and *P*(*B*|*A*)?

f. Are *A* and *B* independent?

**5.17 **For the binomial distribution, do the following:

a. Give the conditions necessary for the binomial distribution to apply to
a random variable.

b. Give the general formula for the probability of *r* successes in *n* trials.

c. Give the probability mass function for *Bi*(10, 0.40).

d. For the distribution in *c*,
determine the probability of no more than four successes.

**5.18 **Sickle cell anemia is a genetic disease that occurs
only if a child inherits two** **recessive
genes. Each child receives one gene from the father and one from the mother. A
person can be characterized as follows: The person can have: (a) two dominant
genes (cannot transmit the disease to a child), (b) one domi-nant and one
recessive gene (has the trait and is therefore a carrier who can pass on the
disease to a child, but does not have the disease), or (c) has both recessive
genes (in which case the person has the disease and is a carrier of the
disease). For each parent there is a 50–50 chance that the child will inher-it
either the dominant or the recessive gene. Calculate the probability of the
child having the disease if:

a. Both parents are carriers

b. One parent is a carrier and the other has two dominant genes

c. One parent is a carrier and the other has the disease

Calculate the probability that the child will be a
carrier if:

d. Both parents are carriers

e. One parent is a carrier and the other has the disease

f. One parent is a carrier and the other has two dominant genes

**5.19 **Under the conditions given for Exercise 5.18,
calculate the probability that** **the
child will have two dominant genes if:

a. One of the parents is a carrier and the other parent has two dominant
genes

b. Both of the parents are carriers

**5.20 **Compute the mean and variance of the binomial
distribution** ***Bi*(*n*,** ***p*). Find the** **arithmetic values for the special case
in which both *n* = 10 and *p* = 1/2.

**5.21** a. Define the probability density and cumulative
probability function for an absolutely continuous random variable.

b. Which of these functions is
analogous to the probability mass function of a discrete random variable?

c. Determine the probability
density function and the cumulative distribution function for a uniform random
variable on the interval [0, 1].

**5.22 **In the example in Section 5.9, consider the
probability that three items have** **mismatched
labels and one of these items is found.

a. Calculate the probability that all three items would pass inspection
and, therefore, there would be two additional ones out of the 84 remaining in
the field.

b. Calculate the probability that exactly one of the two remaining items
with mismatched labels is among the 84 items still in the field. (Hint: Add
to-gether two probabilities, namely the probability that exactly one item is
re-moved at the second stage but none at the third, added to the probability
that exactly one item is removed at the third stage but none at the second).

c. Use the results from (a) and (b) above to calculate the probability that
at least one of the two additional items with mismatched labels is among the 84
remaining in the field.

d. Based on the result in (c), do you think the probability is small enough
not to recall the 84 items for inspection?

__Answers:__

**5.1 **The probability of no females and 4 males is the
same as getting 4 heads in a** **row
tossing a fair coin or (1/2)^{4} = 1/16 = 0.0625. To get one female we
could have the sequence FMMM, which has probability 1/16 also, but there are *C* ^{4}_{1} = 4!/(1! 3!)
= 4 ways of arranging 1 female and 3 males. These 4 mutually exclusive cases
each have probability 1/16. taking the sum the probability is 4/16 = 1/4 =
0.250 for 1 fe-male. For 2 females and 2 males there are *C*^{4}_{2} = 4!/(2! 2!) = 6 ways of getting 2 males
and 2 females. So the probability is 6/16 = 3/8 = 0.375. For 3 females, we
again have 4 ways of getting 3 females and 1 male, so the probability is 0.250
for 3 females. Finally, the probability of getting all 4 females is the same as
the probabil-ity of 0 heads when tossing a fair coin, or 1/16 = 0.0625.

**5.5 **Each of the six faces of a balanced die has the
same chance as any other. So** **the
probability is 1/6 for a single dot, 1/6 for two dots, 1/6 for three dots, 1/6
for four dots, 1/6 for five dots and 1/6 for the face with six dots. These
probabilities also represent the expected proportion of occurrences for each of
these respective faces in 1000 rolls. The expected number of occurrence is
simply *np*, where *n* is the number of rolls and *p*
is the probability of occurrence on an individual roll. Since in this example *n* = 1000 and *p* = 1/6, the expected numberis 166.67 for each of the six faces.

**5.11 **a. This is the number of combinations of seven
objects chosen four at a** **time: *C*(7, 4) = 7!/(4!3!) = 7 × 6 × 5/(3 × 2)
= 35.

b. This is the number of
combinations of six objects chosen four at a time: *C*(6, 4) = 6!/(4!2!) = 6 × 5/2 = 15.

c. This is the number of combinations
of six objects chosen two at a time: *C*(6,2)
= 6!/(2!4!) = 6 × 5/2 = 15. This gives the same result as b.

d. This is the number of
combinations of five objects chosen two at a time: *C*(5, 2) = 5!/(2!3!) = 5 ×
4/2 = 10.

e. 5.11 d is *C*(5, 2), the number of combinations of five objects chosen two at a
time, whereas 5.9 e is *P*(5, 2), the
number of permutations of five objects chosen two at a time. The difference
between permutations and combinations is that in per-mutations the order
matters, whereas in combinations it does not. So if the five ob-jects are
labeled a, b, c, d, and e. There is only one combination for the choice of *a* and *b* but there are two permutations, namely, *ab*, and *ba*. This is true
for each dis-tinct pair. So *P*(5, 2)
is 2*C*(5, 2). In general, *C*(*n*,
*r*) = *P*(*n*, *r*)/*P*(*r*, *r*) = *P*(*n*,
*r*)/*r*! Or *P*(*n*,*
r*) =* r*!* C*(*n*,* r*). In this case,* r *= 2
and* n *= 5.

f. 5.11 b is *C*(6, 4), the number of combinations of six objects chosen four at a
time, whereas 5.9 d is *P*(6, 4), the
number of permutations of six objects chosen four at a time. From e, we saw
that the difference here is the difference between permutations and
combinations. From the general result given in e, we see that *C*(*n*,
*r*) =* P*(*n*,* r*)/*P*(*r*,*
r*) =* P*(*n*,* r*)/*r*! Or*
P*(*n*,* r*) =* r*!* C*(*n*,* r*). In this case,* r *= 4 and* n *= 6. So* P*(6, 4) = 4!* C*(6, 4) = 24* C*(6, 4).

**5.12 **Say that the colors are red, blue, green, and
yellow, denoted R, B, G, and Y,** **respectively.
The possible arrangements are the number of permutations 4! = 4 × 3 × 2 = 24.
They are RBGY, RBYG, RGYB, RGBY, RYGB, RYBG, BRGY, BRYG, BGRY, BGYR, BYRG,
BYGR, GRBY, GRYB, GBYR, GBRY, GYRB, GYBR, YRBG, YRGB, YGRB, YGBR, YBGR, YBRG.

**5.14 **a.** ***C*(4, 2) = 4!/(2!2!) = 4 × 3/2 = 6. This
is the number of combinations of 4** **objects
taken 2 at a time.

*b. P*(5, 3) = 5!/2! = 5 × 4 × 3 = 60. This is the number
of permutations of 5 ob-jects taken 3 at a time.

c. 4!=4×3×2=24.

*d. P*(*A U B*) =* P*(*A*)
+* P*(*B*).

*e. P*(*A **∩** B*) =* P*(*A*)*
P*(*B*).

**5.20 **If** ***X***
**is binomial with parameters** ***n***
**and** ***p*, then the expected value is the sum** **of the expected value on each Bernoulli trial, but if *Y* = 0 with probability 1 – *p* and *Y *= 1 with probability* p*,
then* E*(*Y*) = (1 –* p*)0 +* p*(1) =* p*. Now* E*(*X*) =*
np *since* p *is* *summed *n* times (*n Y*s are added together) the

Var(*X*) = *E*(*X* – *np*)^{2}
= Σ(*k – np*)^{2}*C*(*n*,
*k*)*p ^{k}*(1
–

We now use a few simple tricks.
First recall for any integer *m* > 0

Σ*C*(*m*, *k*)*p ^{k}*(1 –

when the sum is taken from *k* = 0 to *k* = *m*. This is because it
is the sum of probabil-ities for all possible outcomes of a binomial random
variable with parameters *m* and *p*. We will repeatedly use equation (1).
Next, for a binomial random variable* X*,
we* *have seen

*E*(*X*) =* np *=*
*Σ*kC*(*n*,* k*)*p ^{k}*(1 –

We will also exploit equation (2).

Let us consider the third term in the variance
formula first:

Σ*n*^{2}*p*^{2}*C*(*n*, *k*)*p ^{k}*(1 –

using equation (1).

Now consider the second term in the variance
formula:

–Σ2*knpC*(*n*, *k*)*p ^{k}*(1
–

using equation (2).

So the variance equation reduces to

Var(*X*) = Σ*k*^{2}*C*(*n*, *k*)*p ^{k}*(1
–

Now we consider the first term. We use an algebraic
trick:

Σ*k*^{2}*C*(*n*, *k*)*p ^{k}*(1 –

Now by (2) the first term is *np*. Consider the second term:

Σ*k*(*k* – 1)*C*(*n*, *k*)*p ^{k}*(1 –

Notice that in the sum the terms *k* = 0 and *k* = 1 are both 0, so we can take the sum from *k* = 2 to *n*. Let *m* = *n*
– 2 and *j* = *k* – 2. Substituting *j* and
*m* in the equation above we get

The sum on the right side of the
above equation goes from *j* = 0 to *j* = *m*
= *n* – 2. By factoring our *n*(*n*
– 1)*p*^{2} from the summation
we get for the right side,

n(n – 1)p2 = Σ [ (n – 2)! / j!(n – 2 – j)! ] . p^{j}(1
– p)^{n–2–j}

but this sum equals 1 by equation
(1) applied with *m* = *n* – 2 > 0. Note equation (1) holds
trivially for *n* – 2 = 0. So *m* = 0 is also acceptable. So for any *n* ≥ 2, Var(*X*) = *np
*+* n*(*n *– 1)*p*^{2}* *–*
n*^{2}*p*^{2}* *=*
np *+* n*^{2}*p*^{2}* *–* np*^{2}* *–*
n*^{2}*p*^{2}* *=*
np *–* np*^{2}* *=*
np*(1 –* p*) For* n *= 10 and* p *= 1/2, the
mean is 5 and the variance is 10(1/2)(1/2) = 5/2 = 2.5. Note that the* *proof does not include the case *n* = 1, a single Bernoulli trial. In that
case, we com-pute the variance directly, namely, Var(*X*) = *E*[*X* – *E*(*X*)]^{2} and *E*(*X*) = 0(1 – *p*) + 1(*p*) = *p*. So Var(*X*) = *E*(*X* – *p*)^{2}
= (1 – *p*)(0 – *p*)^{2} + (*p*)(1 – *p*)^{2} = (1 – *p*)*p*^{2}
+ *p*(1 – *p*)^{2} = *p*(1 – *p*)(*p*
+ 1 – *p*) = *p*(1 – *p*) = *np*(1 – *p*), since *n* = 1.

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