Problem 10 - Algebra and Trigonometry

Question

Five independent trials of a binomial experiment with probability of success \(p=0.7\) and probability of failure \(q=0.3\) are performed. Find the probability of each event. At least three successes

Step-by-Step Solution

Answer

The probability of at least three successes is approximately 0.83692.

Step 1: Understand the Problem

We are dealing with a binomial distribution with parameters: number of trials \( n = 5 \), probability of success \( p = 0.7 \), and probability of failure \( q = 1 - p = 0.3 \). We are looking for the probability of getting at least three successes in five trials.

Step 2: Define the Cumulative Probability

The probability of at least three successes can be expressed as the sum of probabilities of exactly three, four, and five successes. In mathematical terms, this is \( P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \).

Step 3: Calculate the Probability for Exactly Three Successes

Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Substitute \( n = 5 \), \( k = 3 \), \( p = 0.7 \), so:\[ P(X = 3) = \binom{5}{3} (0.7)^3 (0.3)^{5-3} = 10 \times 0.343 \times 0.09 = 0.3087 \]

Step 4: Calculate the Probability for Exactly Four Successes

Using the binomial formula with \( k = 4 \): \[ P(X = 4) = \binom{5}{4} (0.7)^4 (0.3)^{5-4} = 5 \times 0.2401 \times 0.3 = 0.36015 \]

Step 5: Calculate the Probability for Exactly Five Successes

Using the binomial formula with \( k = 5 \): \[ P(X = 5) = \binom{5}{5} (0.7)^5 (0.3)^{5-5} = 1 \times 0.16807 \times 1 = 0.16807 \]

Step 6: Sum the Probabilities for At Least Three Successes

Add the probabilities for exactly three, four, and five successes:\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) = 0.3087 + 0.36015 + 0.16807 = 0.83692 \]

Key Concepts

Probability of Success

In a binomial experiment, the probability of success is a key concept to grasp. It is denoted by the symbol \( p \) and represents the likelihood of achieving the desired outcome in a single trial. For each attempt, this probability remains constant, which is an essential characteristic of binomial distributions.
Understanding \( p \) helps us determine how likely it is to achieve a certain number of successes over multiple trials. For example, if the probability of success \( p \) is 0.7, as in the given problem, it means there is a 70% chance of success in each trial.

• This value impacts the calculations of individual event probabilities.
• It influences the overall shape of the distribution.

The probability of success, together with the probability of failure \( q \) (where \( q = 1 - p \)), ensures that each trial totals to a certainty of 100%.

Binomial Probability Formula

The binomial probability formula is a crucial tool for calculating the probability of a specific number of successes in a series of independent Bernoulli trials. The formula is expressed as follows: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, the parameters represent:

• \( n \): The total number of trials.
• \( k \): The number of successes we are interested in.
• \( \binom{n}{k} \): The number of combinations of \( n \) items taken \( k \) at a time, calculated by \( \frac{n!}{k!(n-k)!} \).
• \( p \): The probability of success in a single trial.
• \( 1-p \): The probability of failure.

This formula allows us to determine the exact likelihood of achieving exactly \( k \) successes in \( n \) trials. For instance, in the given scenario, you calculated the probability for exactly three, four, and five successes using this method.

Cumulative Probability

Cumulative probability helps us understand the likelihood of achieving a range of outcomes. Instead of calculating the probability for a single event, it sums the probabilities for multiple events.
In the context of the problem, you are interested in the probability of achieving at least three successes in five trials. This is an example of cumulative probability, seeking the probability for \( P(X \geq 3) \).

• To find \( P(X \geq 3) \), add the individual probabilities of achieving exactly three, four, and five successes:
• \( P(X = 3) + P(X = 4) + P(X = 5) \)

This approach is practical because it allows us to understand the likelihood of a sum of different event outcomes within a particular range. Cumulative probability is particularly useful in scenarios where "at least" or "at most" types of questions are involved. This gives a complete picture of the chances for multiple outcomes at once, providing deeper insights into the problem at hand.