2.5.1 problem 57

(a)

Suppose $C_1$ contains $7$ green gummi bears and $8$ red ones, $M_1$ contains $1$ green gummi bear and $2$ red gummi bears, $C_2$ contains $5$ green gummi bears and no red gummi bears, $M_2$ contains $12$ green gummi bears and $5$ red gummi bears.

The proportion of green gummi bears in $C_1$ is $\frac{7}{15}$, which is larger than that of $M_1$, which is $\frac{1}{3}$. The proportion of green gummi bears in $C_2$ is $\frac{5}{5}$, which is larger than that of $M_2$, which is $\frac{12}{17}.$ However, the proportion of green gummi bears in $C_1+C_2$ is $\frac{12}{20}$, which is less than that of $M_1+M_2$, which is $\frac{13}{20}.$

(b)

We can imagine that it is much more difficult to get a green gummi bear out of a jar with subscript $1$ than it is out of a jar with subscript $2$. $C$ jars have a lower overall success rate, because most of their green gummi bears are in $C_1$, which is harder to sample from compared to the jars with subscript $2$.

Let $A$ be the event that a sampled gummi bear is green. Let $B$ be the event that the jar being sampled from is an $M$ jar. Let $C$ be the event that the jar being sampled from has subscript $1$.

Then, by Simpson’s Paradox, $P\left(A\right|B,C)<P(A|B^c,C)$, $P\left(A\right|B,C^c)<P(A|B^c,C^c)$, however, $P\left(A\right|B)>P(A\left|B^c\right).$