Question

Probability of getting at least one head in two coin tosses: ____.

• A. 1/4
• B. 1/2
• C. 3/4
• D. 1

Hint:

Model the number of heads as $X\sim\text{Binomial}(2,1/2)$. Add the probabilities of the two allowed cases, $X=1$ and $X=2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The sample space represents all possible outcomes. "At least one" means one or more.

Step 2: Key Formula or Approach:
Method 1: List favorable outcomes. Method 2: Use the complement rule: $P(\text{At least one}) = 1 - P(\text{None})$.

Step 3: Detailed Explanation:
1. Sample Space ($S$) for 2 coins: $\{HH, HT, TH, TT\}$. Total outcomes = 4. 2. Method 1: Favorable outcomes (At least one H) = $\{HH, HT, TH\}$. Count = 3. \[ P = \frac{3}{4} \] 3. Method 2: The only outcome with no heads is $\{TT\}$. \[ P(\text{None}) = \frac{1}{4} \] \[ P(\text{At least one}) = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 4: Final Answer:
The probability is 3/4.

Answer 2

Concept:

• The number of heads in two fair tosses follows a binomial distribution with $n=2$ and $p=\dfrac12$.
• At least one head means exactly one head or exactly two heads.

Step 1: Find the probability of exactly one head.
$P(X=1)=\binom21\left(\dfrac12\right)^1\left(\dfrac12\right)^1$
$=2\times\dfrac14=\dfrac12$

Step 2: Find the probability of exactly two heads.
$P(X=2)=\binom22\left(\dfrac12\right)^2=1\times\dfrac14=\dfrac14$

Step 3: Add the two disjoint cases.
$P(X\ge1)=P(X=1)+P(X=2)$
$=\dfrac12+\dfrac14=\dfrac34$

Final Answer: $\dfrac34$, option C.