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: Understanding the Question:
Three unbiased coins are tossed simultaneously in a single trial.
We need to calculate the classical probability of the event where we get exactly two tails.

Step 2: Key Formula or Approach:
The probability of an event \(E\) is given by:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in the sample space}} \]
We will list the entire sample space for three tossed coins and then filter the outcomes that contain exactly two tails.

Step 3: Detailed Explanation:

• When a single coin is tossed, there are 2 possible outcomes: Head (H) or Tail (T).

• When three coins are tossed together, the total number of outcomes in the sample space \(S\) is given by:
\[ n(S) = 2 \times 2 \times 2 = 2^3 = 8 \]

• Let us write down all the 8 possible outcomes in the sample space:
\[ S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \]

• Let \(E\) be the event of getting exactly two tails.
- Look through the sample space to find the outcomes containing exactly two 'T's:
- \(HTT\) (one head, two tails)
- \(THT\) (one head, two tails)
- \(TTH\) (one head, two tails)

• The set of favorable outcomes is:
\[ E = \{HTT, THT, TTH\} \]

• Count the number of favorable outcomes:
\[ n(E) = 3 \]

• Calculate the probability of the event \(E\):
\[ P(E) = \frac{n(E)}{n(S)} = \frac{3}{8} \]


Step 4: Final Answer:
The probability of getting exactly two tails is \(\frac{3}{8}\). This corresponds to option (C).