Cards and Independence
Task
One card is selected at random from the following set of 6 cards, each of which has a number and a black or white symbol: $\{\bf 2\triangle, 4\square, 8\blacksquare, 8\blacklozenge, 5\square,
5\blacksquare\}$
 Let $B$ be the event that the selected card has a black symbol, and $F$ be the event that the selected card has a 5. Are the events $B$ and $F$ independent? Justify your answer with appropriate calculations.
 Let $B$ be the event that the selected card has a black symbol, and $E$ be the event that the selected card has an 8. Are the events $B$ and $E$ independent? Justify your answer with appropriate calculations.
IM Commentary
This task lets students explore the concept of independence of events. There are two alternative ways of solving this problem. One way is to use the fact that two events $A$ and $B$ are independent if $P(A \text{ and } B)=P(A) \cdot P(B)$. The alternative approach uses conditional probability and the fact that two events $A$ and $B$ are independent if $P(A  B)=P(A)$, or $P(B  A)=P(B)$. The task can be used to illustrate either of these two approaches.
Solutions
Solution:
Solution 1
To test if the two events $A$ and $B$ are independent we check whether $P(A \text{ and } B)=P(A) \cdot P(B)$.

Out of the six cards, there are three with a black symbol, two with a 5, and only one card has a 5 and a black symbol. Thus we have
$P(B)=\frac36=\frac12$
$P(F)=\frac26=\frac13$
$P(B \text{ and } F)=\frac16$
Since $P(B) \cdot P(F)=\frac12 \cdot \frac13 = \frac16 =P(B \text{ and } F)$, the two events $B$ and $F$ are independent.

Out of the six cards, there are three with a black symbol, and two with an 8, both of them with a black symbol. Thus we have
$P(B)=\frac36=\frac12$
$P(E)=\frac26=\frac13$
$P(B \text{ and } E)=\frac26=\frac13$.
Since $P(B) \cdot P(E)=\frac12 \cdot \frac13 = \frac16 \neq \frac13 = P(B \text{ and } E)$, the two events $B$ and $E$ are not independent.
Solution:
Solution 2
To test if the two events $A$ and $B$ are independent we check whether $P(A  B)=P(A)$.

Out of the six cards, there are two with a 5, so $P(F)=\frac26=\frac13$. Out of the three cards with black symbols, there is only one with a 5, so $P(F  B)=\frac13$.
Since $P(F  B)=P(F)$, the two events $B$ and $F$ are independent.

Out of the six cards, there are two with an 8, so $P(E)=\frac26=\frac13$. Out of the three cards with black symbols, there are two with an 8, so $P(E  B)=\frac23$.
Since $P(E  B) \neq P(E)$, the two events $B$ and $E$ are not independent.