Chapter 3: Problem 19
If the probability of rain on any given day in City \(X\) is 50 percent, what is the probability that it rains on exactly 3 days in a 5 -day period?
Short Answer
Expert verified
P(X = 3) = 0.3125
Step by step solution
01
- Define the problem
We need to find the probability of exactly 3 days of rain in a 5-day period, with each day having a 50% (0.5) chance of rain.
02
- Identify the binomial probability formula
The probability of exactly k successes (raining days) in n trials (days) is given by the binomial formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]is the binomial coefficient, p is the probability of success, n is the number of trials, and k is the number of successes.
03
- Plug values into the formula
Here, n = 5, k = 3, and p = 0.5. Plugging these values into the binomial formula:\[ P(X = 3) = \binom{5}{3} (0.5)^3 (1-0.5)^{5-3} \]
04
- Calculate the binomial coefficient
Calculate the binomial coefficient \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! 2!} = 10 \]
05
- Solve the equation
Now substitute the binomial coefficient and solve the equation:\[ P(X = 3) = 10 \times (0.5)^3 \times (0.5)^2 \]\[ = 10 \times 0.125 \times 0.25 \]\[ = 10 \times 0.03125 \]\[ = 0.3125 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Probability
Binomial probability is the likelihood of a given number of successes in a fixed number of independent trials, where each trial has the same probability of success. For instance, in the context of our rain example, we are trying to find out the probability of having exactly 3 days of rain in a 5-day period. Each day acts as an independent trial, with a probability (\(p\)) of rain being 0.5 (since there's a 50% chance of rain each day). The number of trials (\(n\)) is 5, and the number of successes (\(k\)) is 3.
Probability Formula
The probability formula for binomial distributions helps us calculate the likelihood of a specific number of successes. This formula is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here:
In our specific problem, n = 5, k = 3, and p = 0.5. Plug these values into the formula to find the desired probability.
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here:
- \(\binom{n}{k}\) represents the binomial coefficient, also known as 'n choose k'.
- p is the probability of success in a single trial.
- \(1-p\) is the probability of failure.
- n is the total number of trials.
- k is the number of successes we are interested in.
In our specific problem, n = 5, k = 3, and p = 0.5. Plug these values into the formula to find the desired probability.
Combinatorics
Combinatorics is a branch of mathematics dealing with combinations and permutations of objects. In the context of our probability problem, we use combinatorics to determine the number of ways in which 3 days of rain can occur within 5 days. This is represented by the binomial coefficient:
\[ \binom{5}{3} = \frac{5!}{3! (5-3)!} \]
This coefficient calculates how many different combinations of 3 rainy days out of 5 possible days exist. Here, \(5!\) (5 factorial) represents the product of all positive integers up to 5, (i.e., \(5 \times 4 \times 3 \times 2 \times 1\)). The factorial notation helps account for the positional arrangements of objects. Solving this gives us \(\binom{5}{3} = 10\).
\[ \binom{5}{3} = \frac{5!}{3! (5-3)!} \]
This coefficient calculates how many different combinations of 3 rainy days out of 5 possible days exist. Here, \(5!\) (5 factorial) represents the product of all positive integers up to 5, (i.e., \(5 \times 4 \times 3 \times 2 \times 1\)). The factorial notation helps account for the positional arrangements of objects. Solving this gives us \(\binom{5}{3} = 10\).
Probability Calculations
Now that we've defined the key components, let's perform the calculations step-by-step. First, we calculated the binomial coefficient, which is 10. Next, plugging in the values into our formula, we get:
\[ P(X = 3) = 10 \times (0.5)^3 \times (1-0.5)^{5-3} \]
This simplifies to:
\[ P(X = 3) = 10 \times (0.5)^3 \times (0.5)^2 \]
which then further reduces to:
\[ P(X = 3) = 10 \times 0.125 \times 0.25 \]
Finally, this results in:
\[ P(X = 3) = 10 \times 0.03125 = 0.3125 \]
This means there is a 31.25% chance that it will rain exactly 3 days out of 5 in City X.
\[ P(X = 3) = 10 \times (0.5)^3 \times (1-0.5)^{5-3} \]
This simplifies to:
\[ P(X = 3) = 10 \times (0.5)^3 \times (0.5)^2 \]
which then further reduces to:
\[ P(X = 3) = 10 \times 0.125 \times 0.25 \]
Finally, this results in:
\[ P(X = 3) = 10 \times 0.03125 = 0.3125 \]
This means there is a 31.25% chance that it will rain exactly 3 days out of 5 in City X.