Question

Your experiment consists of tossing an altered coin 3 times successively and recording the side that comes up after each toss. The probability of Head is Pr(H) = 0.4. Calculate the probability of getting exactly two Heads in three successive tosses of the altered coin.

Answer 1

To find the probability of getting exactly two Heads in three coin tosses with Pr(H) = 0.4, calculate the probability for one sequence (0.4 × 0.4 × 0.6) and multiply by the number of such sequences (3), resulting in a probability of 28.8%.

Step-by-step explanation:

The student is interested in understanding the probability of getting exactly two Heads in three successive tosses of an altered coin with a probability of a Head being Pr(H) = 0.4. Utilizing the concept of binomial probability, we can calculate the desired probability.

To achieve exactly two Heads in three tosses, there are three distinct sequences in which this can occur: HHT, HTH, and THH. Since each toss is independent, we can calculate the probability for one such sequence and then multiply by the number of sequences. For one sequence, such as HHT, the probability is 0.4 (for the first Head) × 0.4 (for the second Head) × 0.6 (for the Tail), which equals 0.096. Because there are three such sequences, we multiply 0.096 by 3, yielding a probability of 0.288 or 28.8%.

The calculation steps can be summarized as follows:

1. Calculate the probability of one sequence with two Heads: Pr(HH•) = Pr(H) × Pr(H) × Pr(•), where • represents Tail.
2. Recognize that there are three possible sequences to get two Heads.
3. Multiply the single sequence probability by 3 to get the total probability for exactly two Heads in three tosses.