# Given Information - A biased coin lands heads with probability \( p = .3 \- We are interested in the probability the **first head occurs on the5th toss**. # What Find - Probability that the first head appears on the 5th toss. # Definition or Concept Used This is a case of the **geometric distribution**, which models the number of trials needed for the first success (head). The probability that the first success occurs on the \( n \)-th trial is: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] # Solution To get the first head on the 5th toss: - The first 4 tosses must be tails. - The 5th toss must be a head. Probability of tail in a toss = \( 1 - p = .7 \) So, \[ P(\text{first head on 5th toss}) = (.7)^4 \times .3 \] Calculate \( (.7)^4 \): \[ (.7)^4 = .7 \times .7 \times .7 \times .7 = .2401 \] Now multiply by .3: \[ .2401 \times .3 = .07203 \] # Summary The probability that the first head occurs on the 5th toss is **.07203**.

# Given Information - A biased coin lands heads with probability \( p = 0.3 \). - We need to find the probability that the **first head occurs on the 5th toss**. # What to Find - The probability that the first head appears on the 5th toss. # Definition or Concept Used This problem utilizes the **geometric distribution**, which describes the number of trials until the first success (in this case, a head). The formula for the probability that the first success occurs on the \( n \)-th trial is given by: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] # Solution To find the probability that the first head occurs on the 5th toss, we need to consider the following: 1. The first 4 tosses must result in tails. 2. The 5th toss must result in a head. Calculating the probability of tails for each toss: - The probability of tails in a single toss is \( 1 - p = 0.7 \). Thus, the probability for the first 4 tosses being tails is: \[ P(\text{tails on first 4 tosses}) = (0.7)^4 \] Calculating \( (0.7)^4 \): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Now, we multiply by the probability of getting heads on the 5th toss: \[ P(\text{first head on 5th toss}) = (0.7)^4 \times 0.3 = 0.2401 \times 0.3 \] Calculating the final probability: \[ 0.2401 \times 0.3 = 0.07203 \] # Summary The probability that the first head occurs on the 5th toss is **0.07203**.

# Given Information - A biased coin lands heads with probability \( p = 0.3 \). - We want to find the probability that the **first head occurs on the 5th toss**. # What to Find - The probability that the first head appears on the 5th toss. # Definition or Concept Used This scenario is modeled by the **geometric distribution**, which gives the probability that the first success occurs on the \( n \)-th trial. The formula is: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] Here, \( (1-p) \) is the probability of failure (tails). # Solution To find the probability that the first head occurs on the 5th toss: 1. Calculate the probability of getting tails for the first 4 tosses: - The probability of tails in each toss is \( 1 - p = 0.7 \). Therefore, the probability for the first 4 tosses being tails is: \[ P(\text{tails on first 4 tosses}) = (0.7)^4 \] 2. Now, calculate \( (0.7)^4 \): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] 3. The probability of getting heads on the 5th toss is \( p = 0.3 \). 4. Combine these probabilities to find the probability of the first head on the 5th toss: \[ P(\text{first head on 5th toss}) = (0.7)^4 \times 0.3 = 0.2401 \times 0.3 \] 5. Perform the multiplication: \[ 0.2401 \times 0.3 = 0.07203 \] # Summary The probability that the first head occurs on the 5th toss is **0.07203**.

# Given Information - A biased coin lands heads with probability \( p = 0.3 \). - The objective is to determine the probability that the **first head occurs on the 5th toss**. # What to Find - The probability that the first head appears specifically on the 5th toss. # Definition or Concept Used This problem can be analyzed using the **geometric distribution**, which calculates the likelihood of the first success (head) occurring on the \( n \)-th trial. The relevant formula is: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] where \( (1-p) \) represents the probability of failure (tails). # Solution To compute the probability that the first head occurs on the 5th toss, we follow these steps: - The first four tosses must all be tails, and the fifth toss must be heads. First, calculate the probability of tails in each toss: - The probability of tails in one toss is \( 1 - p = 0.7 \). Thus, the probability of obtaining tails on the first four tosses is: \[ P(\text{tails on first 4 tosses}) = (0.7)^4 \] Next, compute \( (0.7)^4 \): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Now, we consider the probability of obtaining heads on the 5th toss, which is \( p = 0.3 \). Combine these probabilities to find the total probability for the first head on the 5th toss: \[ P(\text{first head on 5th toss}) = (0.7)^4 \times 0.3 = 0.2401 \times 0.3 \] Finally, perform the multiplication: \[ 0.2401 \times 0.3 = 0.07203 \] # Summary The probability that the first head occurs on the 5th toss is **0.07203**.

# Given Information - A biased coin has a probability of landing heads as \( p = 0.3 \). - We need to find the probability that the **first head appears on the 5th toss**. # What to Find - The probability that the first head occurs specifically on the 5th toss. # Definition or Concept Used This situation is modeled by the **geometric distribution**, which describes the number of trials required to achieve the first success (head). The probability formula is: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] Here, \( (1-p) \) represents the chance of failure (tails). # Solution To compute the probability that the first head occurs on the 5th toss, consider the following: - The first four tosses must all result in tails, and the fifth toss must be a head. First, calculate the probability of tails: - The probability of tails for one toss is \( 1 - p = 0.7 \). Thus, the probability that the first four tosses are tails is: \[ P(\text{tails on first 4 tosses}) = (0.7)^4 \] Calculating \( (0.7)^4 \): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Next, the probability of obtaining heads on the 5th toss is \( p = 0.3 \). Now, combine these probabilities to find the total probability of the first head occurring on the 5th toss: \[ P(\text{first head on 5th toss}) = (0.7)^4 \times 0.3 = 0.2401 \times 0.3 \] Perform the multiplication: \[ 0.2401 \times 0.3 = 0.07203 \] # Summary The probability that the first head occurs on the 5th toss is **0.07203**.

# Given Information - A biased coin has a probability \( p = 0.3 \) for landing heads. - We need to determine the probability that the **first head occurs on the 5th toss**. # What to Find - The probability that the first head appears specifically on the 5th toss. # Definition or Concept Used This scenario is modeled using the **geometric distribution**, which expresses the probability of the first success (head) occurring on the \( n \)-th trial. The formula is: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] Here, \( (1-p) \) represents the probability of failure (tails). # Solution To find the probability that the first head appears on the 5th toss, we need to consider the following: - The first four tosses must all be tails. - The fifth toss must be heads. First, we calculate the probability of getting tails in each of the first four tosses: - The probability of tails for one toss is \( 1 - p = 0.7 \). Therefore, the probability of all four tosses resulting in tails is: \[ P(\text{four tails}) = (0.7)^4 \] Next, we compute \( (0.7)^4 \): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Now, we consider the probability of getting heads on the 5th toss, which is \( p = 0.3 \). We combine these probabilities to find the total probability of the first head occurring on the 5th toss: \[ P(\text{first head on 5th toss}) = (0.7)^4 \times 0.3 = 0.2401 \times 0.3 \] Finally, we perform the multiplication: \[ 0.2401 \times 0.3 = 0.07203 \] # Summary The probability that the first head occurs on the 5th toss is **0.07203**.

# Given Information - A biased coin has a probability of landing heads \( p = 0.3 \). - We aim to find the probability that the **first head occurs on the 5th toss**. # What to Find - The probability that the first head appears specifically on the 5th toss. # Definition or Concept Used This situation is modeled by the **geometric distribution**, which is useful for determining the probability of the first success (head) occurring on the \( n \)-th trial. The formula is: \[ P(\text{first success on } n\text{-th trial}) = (1-p)^{n-1} p \] In this context, \( (1-p) \) represents the probability of failure (tails). # Solution To find the probability that the first head occurs on the 5th toss, we need to consider the following conditions: - The first four tosses must all result in tails. - The fifth toss must result in heads. First, we calculate the probability of getting tails for each of the first four tosses: - The probability of tails in one toss is \( 1 - p = 0.7 \). Thus, the probability of getting tails for the first four tosses is: \[ P(\text{tails on first 4 tosses}) = (0.7)^4 \] Next, we compute \( (0.7)^4 \): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Now, consider the probability of obtaining heads on the 5th toss, which is \( p = 0.3 \). We combine these probabilities to find the total probability that the first head occurs on the 5th toss: \[ P(\text{first head on 5th toss}) = (0.7)^4 \times 0.3 = 0.2401 \times 0.3 \] Finally, we perform the multiplication: \[ 0.2401 \times 0.3 = 0.07203 \] # Summary The probability that the first head occurs on the 5th toss is **0.07203**.