Let's analyze and solve each part of this probability problem based on the dartboard shown:

1. Identify white and grey slices: The dartboard has six slices labeled 1 through 6. It looks like:
   • White slices: 1, 2, 4
   • Grey slices: 3, 5, 6

Part (a)

What is the probability that the dart lands on a white slice?

• Total slices = 6
• White slices = 3 (slices 1, 2, and 4)

The probability of landing on a white slice is: $\frac{\text{Number of white slices}}{\text{Total slices}} = \frac{3}{6} = \frac{1}{2}$

Part (b)

What is the probability that the dart lands on a white slice, given that it lands on an odd-numbered slice?

• Odd-numbered slices = 1, 3, 5 (total of 3 odd-numbered slices)
• Among these, the white odd-numbered slice is only slice 1.

The probability that it lands on a white slice, given it lands on an odd-numbered slice, is: $\frac{\text{Number of white odd-numbered slices}}{\text{Total odd-numbered slices}} = \frac{1}{3}$

Would you like further details on the solution steps?

Related Questions:

1. What would the probability be if there were 8 slices instead of 6?
2. How would the probability change if slice 4 was grey?
3. What if the dartboard had an uneven number of white and grey slices?
4. How do probabilities differ between mutually exclusive and non-mutually exclusive events?
5. How does conditional probability affect the outcome in part (b)?

Tip: In probability, always verify the total number of favorable outcomes versus possible outcomes to ensure accurate calculations.