Question

a coin is tossed four times and the number of heads that appear is counted. fill in the following table listing the probabilities of obtaining various numbers of heads. what do you observe? are all of these outcomes equally likely?
number of heads \t0 \t1 \t2 \t3 \t4
probability \t \t \t \t \t
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Explanation:

Step1: Determine total outcomes

When a coin is tossed 4 times, each toss has 2 outcomes (H or T). So total outcomes \( n = 2^4 = 16 \).

Step2: Calculate combinations for each heads

• For 0 heads (all tails): \( C(4,0)=\frac{4!}{0!(4 - 0)!}=1 \). Probability: \( \frac{1}{16} \).
• For 1 head: \( C(4,1)=\frac{4!}{1!(4 - 1)!}=4 \). Probability: \( \frac{4}{16}=\frac{1}{4} \).
• For 2 heads: \( C(4,2)=\frac{4!}{2!(4 - 2)!}=6 \). Probability: \( \frac{6}{16}=\frac{3}{8} \).
• For 3 heads: \( C(4,3)=\frac{4!}{3!(4 - 3)!}=4 \). Probability: \( \frac{4}{16}=\frac{1}{4} \).
• For 4 heads: \( C(4,4)=\frac{4!}{4!(4 - 4)!}=1 \). Probability: \( \frac{1}{16} \).

Answer:

Number of Heads	Probability
1	\( \frac{1}{4} \)
2	\( \frac{3}{8} \)
3	\( \frac{1}{4} \)
4	\( \frac{1}{16} \)

Observation: The probabilities are not equal (e.g., \( \frac{1}{16}
eq\frac{3}{8} \)), so outcomes (number of heads) are not equally likely.