Using a two-way table to find club member probabilities

Key takeaway: This problem tests how to read a contingency table, use conditional probability, and apply multiplication rules for dependent selections.

Key idea

The table gives the full breakdown of the 60 club members by age group and skill level:

• Junior Beginners = 35
• Junior Advanced = 5
• Senior Beginners = 7
• Senior Advanced = 13

So the total number of Beginners is:

$35+7=42$

and the total number of Juniors is:

$35+5=40$

That is enough to answer parts (a) and (b) directly.

Part (a): Probability a randomly chosen member is a Beginner

For a random choice from the whole club, the probability is

$P(\text{Beginner})=\frac{42}{60}=\frac{7}{10}.$

So the answer is $\frac{7}{10}$.

Part (b): Probability a Junior member is a Beginner

Now the sample space is only the Junior row. There are 40 Junior members in total, and 35 of them are Beginners.

So

$P(\text{Beginner} \mid \text{Junior})=\frac{35}{40}=\frac{7}{8}.$

So the answer is $\frac{7}{8}$.

Part (c): Two members chosen at random, both Juniors and exactly one Beginner

This is a two-step probability problem with selection without replacement.

We want both chosen members to be Juniors, and among those two Juniors, exactly one must be a Beginner. The Junior group has 35 Beginners and 5 Advanced members.

A clean way is to count ordered cases:

1. First Junior is Beginner, second Junior is Advanced:

$\frac{35}{60}\times\frac{5}{59}$

2. First Junior is Advanced, second Junior is Beginner:

$\frac{5}{60}\times\frac{35}{59}$

Add the two cases:

=2\cdot\frac{175}{3540} =\frac{350}{3540} =\frac{35}{354}.$$ So the probability is **$\frac{35}{354}$**. ## Why this works The phrase “exactly one is a Beginner” means one Junior Beginner and one Junior Advanced. Because the members are chosen randomly without replacement, the second probability changes after the first choice. That is why the product rule must use 59 remaining members in the second step. A common shortcut is to use combinations: $$\frac{\binom{35}{1}\binom{5}{1}}{\binom{40}{2}}$$ This gives the same answer because it counts unordered pairs among Juniors only. ## Final answers - **(a)** $\frac{7}{10}$ - **(b)** $\frac{7}{8}$ - **(c)** $\frac{35}{354}$ --- **Pitfalls the pros know** 👇 A frequent mistake is to use the total club size of 60 in part (b). The question narrows the sample space to Junior members only, so the denominator must be 40, not 60. Another trap in part (c) is to forget that “exactly one is a Beginner” does not mean choosing one Beginner and one Advanced from the whole club without checking they are both Juniors. The category restriction comes first; only the Junior row is relevant. **What if the problem changes?** If the question changed to “A member is chosen at random. Find the probability that the member is a Junior Beginner,” then the event would be a single table cell, so the answer would be $35/60$. If it changed to “Two members are chosen without replacement; find the probability both are Juniors,” then you would use $40/60\times39/59$. If it asked for “exactly one Beginner” among any two members from the full club, you would need all mixed pairs from the entire table, not just the Junior row. `Tags`: conditional probability, contingency table, sampling without replacement