HW 3 Solutions

p. 166 #3

3a
    i) P(Female and slack work) = .101
    ii)P(slack work | female) = .101/.383 = .2637

3b
    i) P(Male and slack) = .214
    ii) P(slack | male) = .214/.617 = .3468
    iii) P(slack) = .315

3c
    i) P(abolish | females) = .065/.383=.1697
    ii) P(abolish | males) = .098/.617= .1588
    iii) .163

Note: these solutions were based on Table 4.4.2 and due to rounding errors the numbers differ slightly from those in the back of the book (which were based on TAble 4.4.1)

p. 188, 10

a) (1/4)*(1/10) = 1/40 = .025 of schizophrenics kill selves.
b)

	Schizophrenic	Not Schizophrenic	Total
Homeless	(1/3)*.008 = .002666667	.08-.026667=.053333	.088
Homed	.01 - .0026666667=.007333	1-.002666667-.0533333-.007333=.9366667	.944
	.01	.98999997	1 (except for rounding error)

P(homeless|schizophrenic) = .00266667/.01 = .266667

NOte that asking how many people are both homeless and schizophrenic (the first cell in the table) is different from asking what proportion of the schizophrenics are homeless.

c) If genes had no effect, then we woud expect to see about the same proportion of schizophrenics among families with 1 schizophrenic parent as we see among families with 2.  In fact, we see that the child is schizophrenic much more often in families with both parents schizophrenic (.40) then when only one parent is (.10).  So this argues tht genes play a role.  On the other hand, if you take a child of schizophrenic parents out of their home and put in a new environment, it doesn't affect the probabilities thjt the child will be schizophrenic.  So environment seems to hve little effect.

d) P(both parents are | child is) = P(child is and both parents are)/P(child is)
The numerator is P(Child is and both parents are) = P(child is | both parents are) *P(both parents are) = .40 * p
where p is unknown probability that both parents are schizophrenic.

The denominator is .01 based on the idea tht a randomly selected person from the population will be schizophrenic.  So we hve
P(both parents are | child is) = .4*p/.01

e) P(Both parents are) = P(father is AND mother is)  if these are independent, as the problem says to assume = P(father is) * P(mother is)
and, agian, assuming thta there is a 1% chance a randomly selected person is schizophrenic gives this as .01*.01 = .0001

So P(both parents are | child is) = .4*.0001/.01 = .004

f) Frankly, I'm not sure what's being aksed here.  However, in the lat problem we assumed tht 1% of the population was schizophrenic, and we needed this to calculated the probability that a randomly selected child was schizophrenic.  This suggests, though, thta there would be fewer schizophrenics among children (because they hvaen't developed it yet), and this means thta P(child is) should be a number less than .01.  The result would be to make the answer to (e) larger.