A fair die is tossed once. Find the probability of getting an even number or a composite number. 2. A one-way ANOVA was conducted to compare the mean scores of students in Biostatistics, Chemistry, an

3.
In a single toss of a die, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is 6.

Step 1: Identify even numbers ($E$).

$$E = \{2, 4, 6\}$$

Step 2: Identify composite numbers ($C$).
A composite number is a positive integer that has at least one divisor other than 1 and itself.

$$C = \{4, 6\}$$

Step 3: Identify numbers that are even OR composite ($E \cup C$).

$$E \cup C = \{2, 4, 6\}$$

The number of favorable outcomes is 3.

Step 4: Calculate the probability.

$$P(\text{even or composite}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$$

The probability of obtaining an even number or a composite number is $\boxed*{\frac{1}{2} \text{ or } 0.5}$.

4.
To calculate the F-ratio, we perform a one-way ANOVA.
Given data for 11 students across 3 subjects:
Biostatistics (B): 1, 1, 0, 0, 1, 3, 0, 2, 7, 0, 0
Chemistry (C): 0, 2, 1, 1, 2, 6, 7, 3, 5, 4, 3
Biology (L): 1, 7, 5, 1, 2, 0, 4, 1, 2, 3, 0

Number of groups ($k$) = 3.
Number of observations per group ($n_j$) = 11.
Total number of observations ($N$) = $3 \times 11 = 33$.

Step 1: Calculate sums and means for each group.

$$\sum B = 15 \quad \bar{B} = \frac{15}{11} \approx 1.3636$$ $$$$