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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.
Cards and Independence
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.
Comments
Log in to commentJeff Meyer says:
over 4 years(a) is somewhat vague in its writing, in that it's not clear if B and F are conducted sequentially. That is, it's not clear if there is replacement of the cards in the deck, which definitely changes P(B) and P(F). Also  why use "B" and "F" as the function names when "A" and "B" are used in the formal definition of independence? Suggest either using "A" and "B," as in the definition, or two totally different terms.
Also  writing the steps sequentially, as in P(B)=3/6=1/2, P(F)=2/6=1/3, and P(B and F)=1/6 seems odd visually, as opposed to:
roxypeck says:
over 4 yearsThanks for the comment Jeff. I revised the wording of the task to make it clear that only one card is being selected, and reformatted the solution as you suggested. The author used the symbols B and F to be decriptive of the events (B for black and F for five), and since this is something that students will encounter in practice, I didn't change those.