Question

The probability of getting head in a toss of a biased coin is ⅔ . Let the coin be tossed three times independently. Then the probability of getting head in the first two tosses and tail in the final toss is

• A. 45409
• B. 45299
• C. 45349
• D. 23/27

Answer 1

The Correct Option is A

To find the probability of getting heads in the first two tosses and tails in the final toss for a biased coin where the probability of heads is \(\frac{2}{3}\), we can follow these steps:

1. Given: Probability of heads (H) = \(\frac{2}{3}\). Therefore, the probability of tails (T) = 1 - \(\frac{2}{3}\) = \(\frac{1}{3}\).
2. The events are independent, so we can multiply the probabilities for individual outcomes:
3. Probability of getting heads on the first toss = \(\frac{2}{3}\)
4. Probability of getting heads on the second toss = \(\frac{2}{3}\)
5. Probability of getting tails on the third toss = \(\frac{1}{3}\)
6. Thus, the combined probability for getting Heads, Heads, and Tails \((HHT)\) is:

\(\text{P(HHT)} = \text{P(H)} \times \text{P(H)} \times \text{P(T)} = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\)

Computing this:

\(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}\)

Thus, the probability of getting heads on the first two tosses and tails on the final toss is \(\frac{4}{27}\).