|3.12||The probability of the intersection is 0.02, the product of the probabilities is 0.01. The events are therefore not independent|
|3.15||P[A | B ] is the P[A intersected B]/P[A] = 0.02/0.10=0.20|
|3.16||Let B=A2 and A=A1 and Ac the complement of A. The conditional probability is P[B | Ac] = P[ B intersect Ac]/P[ A c] = P[B]-P[B intersect A]/(1-P[A]) = 0.10-0.02/(1-0.10) = 0.089|
|3.17||Let A be the event that a 77 year old man is affected and B be the event that a 76 year old man is affected and C the event that a 82 year old man is affected. By assumption, all these events are independent. The probability that three people taken from each group are affected is P[A intersect B intersect C] = 8.8/100000|
|3.24||Let A be the event that a woman in the age group 75-79 is affected and B be the event that a man in that age group is affected. We have P[B | A] = 0.0015/0.023 = 0.031. The unconditional probability P[B] = 0.049 is slightly larger indicating a correlation.|
|3.25||Let A be the event that a woman in the age group 75-79 is affected and B be the event that a man in that age group is affected. We have P[A | B] = 0.0015/0.049 = 0.031. The unconditional probability P[A] = 0.023 is slightly smaller. A common environmental factor associated with Alzheimer's desease would have the effect that the conditional probability is heigher then the unconditional probability.|
|3.28||The expected number of cases with Alzheimer's disease is 1000 0.061 = 61|
|3.76||Let A be the event that the difference of the automated Diastolic blood pressure has increased by bigger or equal than 10mmHG and B the event that the difference of the manual Diastolic blood pressure is bigger or equal than 10mmHG. Let N be the number of measurements. From Table 3.9, we see that the probability of A intersected with B is 6/N and the probability of B is P[B]=13/N and the probability of A is P[A]=21/N. The sensitivity is P[A | B] = P[ A intersect B ]/P[B] = (6/N)/(13/N) = 6/13 = 0.462.|
|3.77||The specificity of the automated Diastolic blood pressure measurements is P[Ac | Bc] = (51/N)/(66/N) = 51/66 = 0.773.|
|3.78|| With the notation from above the predictive values
positive (PV+) and predictive values negative (PV-) are |
PV+ = P[ B | A] = P[B intersect A]/P[A]=6/21 = 0.286
PV- = P[ Bc | Ac] = 51/58 = 0.879
|3.83||The probabilities in the prevalence columns are P[Yi]. The conditional probabilities P[Xi|Yj] are given in the matrix of Table 3.11. Look up the entry in the 4'th row and 5'th column: P[X5|Y4] = 0.10.|
|3.89||Use Bayes rule to determine P[Y2 | X1 intersect X7] = 0.187|
While commen sense tells the right answer 3=0.2*15 immediatly,
it is useful to formulate the problem precicely using the language
of probability theory. |
Let D be the group of all grade school students and A be the population of grade school students who develop influenca. Assume D contains N students. The probability of A, P[A] is known to be 0.2. Let B be the members of the class under consideration and C is the subset of students in B which develop influenca.
We have C = A intersect B. It is an assumption that B is independent of A. Therefore, we have P[C] = P[A intersect B] = P[A] P[B] = 0.2 * P[B] = 0.2 * 15/N. The expected number of students is N P[C ] = 0.2*15 = 3.
|4.14||X is the Binominal distributed random variable giving the number of inner-city newbotrns with HIV positive test results. The probability p to be HIV positive is 30/3741. The probability 1-p to be HIV negative is 3711/3741. The Binominal formula (which assumes that each birth event is independent from the others) gives P[X=5] = 3741!/(5! 3736!) p5 (1-p)3711 = 0.1573.|
|4.15|| P[X > 4] = 1- P[X < 5].
This probability can be determined easily with Mathematica
a guide how to use Mathematica at Harvard from any computer. The
computer does not need to have Mathematica installed but have access to
the Web). |
<< Statistics`DiscreteDistributions` N[1- CDF[BinomialDistribution[500, 30/3741], 4]]which gives 0.373.
|4.20||As in 4.14. The Binominal formula gives P[X3=5] = 0.0004|
|Last update, 12/18/2000|