Question

A coin is based so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

• A. \(\frac{2}{9}\)
• B. \(\frac{1}{9}\)
• C. \(\frac{2}{27}\)
• D. \(\frac{1}{27}\)

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the probability of getting two tails and one head when a biased coin is tossed three times.

The problem states that the probability of getting a head (\(P(H)\)) is twice the probability of getting a tail (\(P(T)\)). Let's denote the probability of getting a tail as \(p\). Therefore, the probability of getting a head will be \(2p\).

Since the total probability must equal 1, we have:

\(p + 2p = 1\)

Simplifying gives:

\(3p = 1 \Rightarrow p = \frac{1}{3}\)

Thus, \(P(T) = \frac{1}{3}\) and \(P(H) = \frac{2}{3}\).

Now, we wish to find the probability of getting exactly two tails and one head in three tosses. The number of favorable sequences for two tails and one head is given by the combination:

\(C(3,2) = 3\)

This reflects the sequences: TTH, THT, HTT.

The probability for each of these sequences is calculated as:

\(P(\text{sequence}) = (P(T))^2 \cdot (P(H)) = \left(\frac{1}{3}\right)^2 \cdot \frac{2}{3} = \frac{1}{9} \cdot \frac{2}{3} = \frac{2}{27}\)

Since there are 3 sequences, the total probability is:

\(3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9}\)

Therefore, the probability of getting two tails and one head is \(\frac{2}{9}\), which matches the correct answer option.

Answer 2

Define probabilities for head and tail. Let the probability of getting a tail be \( \frac{1}{3} \). Since a head is twice as likely to occur as a tail, the probability of getting a head is:

\[ \text{Probability of head} = 2 \times \frac{1}{3} = \frac{2}{3}. \]

Calculate the probability of getting two tails and one head. The scenario "two tails and one head" can happen in three possible orders: \(\{ \text{TTH, THT, HTT} \}\). The probability of each specific order is:

\[ \left( \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} \right). \]

Thus, the probability of getting exactly two tails and one head is:

\[ \left( \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} \right) \times 3 \] \[ = \frac{2}{27} \times 3 = \frac{2}{9}. \]

Therefore, the answer is:

\[ \frac{2}{9}. \]