Probability: Venn Diagrams and Two-Way Tables - Variation Theory - Free Printable

Let's solve the "Your Turn" section step by step using the provided two-way table. The total number of students is 120.

Given Table:

| | Left handed | Right handed | Total |
|---------------|-------------|--------------|-------|
| French | 20 | 40 | 60 |
| Spanish | 18 | 42 | 60 |
| Total | 38 | 82 | 120 |

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Question 1: Find the probability of a student taking French or Spanish

We are asked to find $ P(\text{French or Spanish}) $.

Since every student takes either French or Spanish, and there are no students outside these two subjects (total = 60 + 60 = 120), this means all students take either French or Spanish.

So:
$$
P(\text{French or Spanish}) = \frac{\text{Total students}}{\text{Total students}} = \frac{120}{120} = 1
$$

✔ Answer: 1 (or 100%)

> Note: Even if we used the formula:
> $$
> P(A \cup B) = P(A) + P(B) - P(A \cap B)
> $$
> But since every student takes one of them, and assuming no student takes both (based on the totals adding up to 120), it's safe to say they are mutually exclusive in this context.

But let’s check:
- French: 60 students
- Spanish: 60 students
- Total: 120 → So no overlap, i.e., no student takes both.

So yes, mutually exclusive.

Thus:
$$
P(\text{French or Spanish}) = P(\text{French}) + P(\text{Spanish}) = \frac{60}{120} + \frac{60}{120} = 0.5 + 0.5 = 1
$$

✔ Answer: 1

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Question 2: Find the probability of a student taking French and is right handed

This is asking for:
$$
P(\text{French and Right-handed}) = \frac{\text{Number of students who take French and are right-handed}}{\text{Total students}}
$$

From the table: French and Right-handed = 40

So:
$$
P = \frac{40}{120} = \frac{1}{3} \approx 0.3333
$$

✔ Answer: $ \frac{1}{3} $ or approximately 0.333

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Question 3: Given that a student is right handed, what is the probability they take French?

This is a conditional probability:
$$
P(\text{French} \mid \text{Right-handed}) = \frac{P(\text{French and Right-handed})}{P(\text{Right-handed})}
$$

We already know:
- $ P(\text{French and Right-handed}) = \frac{40}{120} $
- $ P(\text{Right-handed}) = \frac{82}{120} $

So:
$$
P(\text{French} \mid \text{Right-handed}) = \frac{40/120}{82/120} = \frac{40}{82} = \frac{20}{41} \approx 0.4878
$$

✔ Answer: $ \frac{20}{41} $ or approximately 0.488

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Question 4: Given that a student is right handed, what is the probability they take Spanish?

Similarly:
$$
P(\text{Spanish} \mid \text{Right-handed}) = \frac{P(\text{Spanish and Right-handed})}{P(\text{Right-handed})}
$$

From the table: Spanish and Right-handed = 42
Right-handed total = 82

So:
$$
P = \frac{42}{82} = \frac{21}{41} \approx 0.5122
$$

✔ Answer: $ \frac{21}{41} $ or approximately 0.512

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✔ Final Answers:

1) $ P(\text{French or Spanish}) = 1 $
2) $ P(\text{French and Right-handed}) = \frac{1}{3} $
3) $ P(\text{French} \mid \text{Right-handed}) = \frac{20}{41} $
4) $ P(\text{Spanish} \mid \text{Right-handed}) = \frac{21}{41} $

Let me know if you'd like a visual explanation or how this compares to the worked example!