Question

Three coins are tossed together. The probability of getting exactly two tails is

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

Hint:

You can use the binomial formula \(P(X = k) = \binom{n}{k} p^k q^{n-k}\) where \(n=3\), \(k=2\), and \(p = q = \frac{1}{2}\):
\[ P(X = 2) = \binom{3}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{3-2} = 3 \times \frac{1}{8} = \frac{3}{8} \]
This binomial approach is highly useful when the number of coins is larger.

Answer 1

The Correct Option is C

Step 1: List the sample space.
When three coins are tossed together, each coin shows either Heads (H) or Tails (T). The total sample space is: $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Step 2: Count total outcomes.
Total number of outcomes $= 2^3 = 8$
Step 3: Identify favourable outcomes.
We need exactly two tails. The outcomes with exactly 2 tails are: HTT, THT, TTH. So number of favourable outcomes = 3.
Step 4: Apply the probability formula.
\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]
Step 5: Substitute values.
\[ P(\text{exactly two tails}) = \frac{3}{8} \]
Step 6: Verify the answer matches the options.
Option 3 gives $\frac{3}{8}$, which matches our calculation.
\[ \boxed{\dfrac{3}{8}} \]