Question

A fair coin is tossed five times. Determine the probability that: (a) It turns up tails every time. (b) It turns up heads at most three times. (c) It turns up heads twice in a row exactly one time.

Short answer

(a) 0.03125, (b) 0.8125, (c) Requires detailed case enumeration.

Step-by-step solution

Understanding the Coin Toss Probability

A fair coin has two outcomes: heads (H) and tails (T), each with a probability of 0.5. When it is tossed multiple times, the outcomes for each toss are independent.

Probability of Five Tails in a Row

To calculate the probability of getting tails in all five tosses, use the formula for independent events: \[ P(TTTTT) = (0.5)^5 = 0.03125 \] Thus, the probability of getting tails every time in five tosses is 0.03125.

Calculating 'At Most Three Heads'

To find the probability of getting at most three heads, calculate the probabilities for 0, 1, 2, and 3 heads and sum them up using the binomial formula: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 5 \) (tosses), \( k \) (number of heads), \( p = 0.5 \) (probability of heads). Calculate: - \( P(0) = \binom{5}{0} (0.5)^0 (0.5)^5 = 0.03125 \) - \( P(1) = \binom{5}{1} (0.5)^1 (0.5)^4 = 0.15625 \) - \( P(2) = \binom{5}{2} (0.5)^2 (0.5)^3 = 0.3125 \) - \( P(3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 0.3125 \) Sum: \[ P(\text{at most 3 heads}) = 0.03125 + 0.15625 + 0.3125 + 0.3125 = 0.8125 \]

Probability of Exactly One Double Heads Sequence

To find the probability of exactly one occurrence of two consecutive heads in five tosses, we analyze valid sequences: Double heads can only occur once, leaving room for sequences like HTC/HT, HTHC, THTH, etc. This scenario is complex due to dependency on the rest of the sequence. Construct sequences like HHTTT, THTHT, etc., and calculate their probability. Each such sequence must satisfy the condition of only one HH occurrence. Due to complexity, use enumeration or Monte Carlo simulations for exact probability calculation.

Key concepts

Independent Events

When we talk about independent events in probability theory, we mean events where the outcome of one event does not affect the outcome of another. Each coin toss is an independent event because the result of one toss (heads or tails) does not influence the result of the next toss. This is a key concept, especially in scenarios where we conduct repeated experiments, like tossing a coin multiple times.

For example, when you toss a fair coin, the likelihood of landing heads or tails is always 50% for each toss. This probability remains constant, regardless of previous outcomes. This independence allows us to calculate the probability of a series of outcomes by multiplying the probabilities of individual outcomes.

In our exercise, this principle helps us calculate the probability of getting all tails in five tosses as \( (0.5)^5 \), which equals 0.03125. Independent events simplify complex probability problems and are crucial in understanding binomial probability and combinatorics.

Binomial Probability

Binomial probability is a specific probability distribution that is applicable when there are two possible outcomes, like flipping a coin which results in heads or tails. This distribution helps in calculating the probabilities of obtaining a certain number of successes (e.g., heads) in a specific number of trials (e.g., 5 coin tosses).

The formula used for computing binomial probability is \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:

• \( n \) is the number of trials
• \( k \) is the number of successes
• \( p \) is the probability of success on an individual trial

For our task, we use this formula to find the probability of getting at most three heads in five tosses.

This involves summing the probabilities of getting zero heads, one head, two heads, and three heads. These calculated probabilities are then added together, resulting in a total probability of 0.8125 for getting at most three heads.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations and permutations of objects. It is an essential tool for determining the number of possible outcomes in probability problems. When addressing how likely a specific sequence occurs, combinatorics helps us consider different arrangements and selections.

In our exercise, combinatorics becomes evident in discovering sequences like two consecutive heads appearing exactly once. By creating and analyzing possible sequences, such as HHTTT and THTHT, combinatorial techniques allow us to count these configurations.

The complexity increases as restrictions are added, for example, having exactly one sequence of two heads in a row. Sometimes, when calculations are too intricate for manual counting, techniques like enumeration or simulations provide assistance.

Combinatorics is vital for examining varied configurations and calculating their probabilities accurately, applying these strategies broadens normal probability analysis into more complicated sequence probabilities.