Question

Three coins are tossed together. The probability of getting exactly one head, is :

• A. \(\frac{1}{8}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{3}{8}\)

Hint:

Using binomial expansion or combinations can speed up the calculation.
The number of ways to choose exactly \(1\) head from \(3\) tosses is given by \(\binom{3}{1} = 3\).
Since the total outcomes are \(2^3 = 8\), the probability is directly \(\frac{3}{8}\).

Answer 1

The Correct Option is D

Step 1: Use the binomial probability formula instead of listing the sample space.
Tossing three fair coins is a binomial situation with $n=3$ trials, where each trial (a coin) has probability $p=\dfrac{1}{2}$ of showing heads. The probability of getting exactly $k$ heads is:
\[ P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} \]

Step 2: Substitute $n=3$, $k=1$, $p=\dfrac{1}{2}$.
\[ P(X=1) = \binom{3}{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{2} \]

Step 3: Simplify.
\[ \binom{3}{1} = 3 \]
\[ P(X=1) = 3 \times \frac{1}{2} \times \frac{1}{4} = \frac{3}{8} \]

Final Answer:
The probability of getting exactly one head is $\dfrac{3}{8}$, matching option (D).
\[ \boxed{P = \frac{3}{8}} \]