Problem 33 - Algebra 2

Question

A family is planning a four-day campout. The weather forecast shows a 20\(\%\) chance of rain for each of the four days. Find the probability, to the nearest hundredth, that there will be no rain on at least three of the four days.

Step-by-Step Solution

Answer

The probability that there will be no rain on at least three of the four days is approximately \(0.82\) or \(82\%\).

Step 1: Determine the complementary event

Firstly, we need to find the probability complementary to the given chance of rain, this is the probability of 'no rain'. Since the total probability of all possible events is 1 (or 100%), the probability of no rain (\(P_{nr}\)) is \(1 - 0.20 = 0.80\) or 80%.

Step 2: Apply the Binomial Theorem

Now that we have the probability for no rain (\(P_{nr} = 0.8\)), we can apply the Binomial Theorem. We want at least 3 days of no rain out of 4, so we sum the probabilities for 3 and 4 days of no rain: \[ P = {4 \choose 3} (P_{nr})^3 (1-P_{nr}) + {4 \choose 4} (P_{nr})^4 \] Where \({4 \choose n}\) are binomial coefficients calculated as \({4 \choose n}=\frac{4!}{n!(4-n)!}\) where \(n!\) stands for factorials.

Step 3: Calculate the probability

We find the binomial coefficients: \({4 \choose 3}=4\) and \({4 \choose 4}=1\). Then, substitute the \(P_{nr} = 0.8\) into the equation and calculate: \[P = 4*(0.8)^3 (0.2) + 1*(0.8)^4 = 4*(0.4096) + 0.4096 = 2.048\] Since probability can't be more than 1, we've possibly made an mistake. Let's check the calculations for \(P_{nr}\), as it's easy to confuse the percentages with fractions. When we revise, we remember that \(P_{nr}\) is already in decimal form as \(0.8\), not as \(80\). To correct the error, we calculate the values again: \[P = 4*(0.8)^3 * 0.2 + (0.8)^4 = 0.8192 + 0.4096 = 1.2288\] The probability is still greater than 1, which indicates there's still something wrong with our calculations. When someone rechecks, they realize we've made an error when calculating \((0.8)^3 * 0.2\). It should be 0.1024, not 0.8192. So we have to correct that calculation. In case we want to avoid such mistakes, it might be better to use calculator. Using calculator, we get correct answer: \[P = 4*(0.8)^3 * 0.2 + (0.8)^4 = 0.4096 + 0.4096 = 0.8192\]

Step 4: Round off the answer

Finally, round the final answer to the nearest hundredth as asked in the exercise. Therefore, the probability that there will be no rain on at least three of the four days is \(0.82\) or \(82\%\).

Key Concepts

Probability

Probability helps us understand the likelihood of different outcomes. For example, when a weather forecast shows a 20% chance of rain, this means there's a likelihood of 0.2 for rain on any given day.

There are different types of probabilities, such as experimental and theoretical probability. In this exercise, we are using theoretical probability, based on known data (the 20% chance of rain).

By understanding probability, we can better predict events in uncertain situations. Here, we used this concept to assess various weather scenarios during a campout.

Binomial Theorem

The Binomial Theorem is a key tool for calculating probabilities, especially when dealing with repeated independent events, like several days of weather forecasts.

In this exercise, we deal with four potential rainy days, each with its own probability, independent of the others. We determine outcomes such as at least three days without rain.

The formula used:

• \[{n \choose k} (p)^k (1-p)^{n-k}\]

where

• \(n\) is the total number of trials (4 days here)
• \(k\) is the number of successful outcomes (like 3 days with no rain)
• \(p\) is the probability of a single success (0.8 for no rain)

This theorem makes it easy to manage these probabilities.

Complementary Events

Complementary events are pairs of outcomes that cover all possibilities together. For weather forecasts, knowing there's a 20% chance of rain automatically means there's an 80% chance of no rain, because together they must add up to 100%.

In probability terms, if event A is one outcome, its complement (A') is the opposite. For any event, \[P(A) + P(A') = 1\] holds true.

This concept helps simplify calculations by focusing directly on what you're interested in (like no rain), instead of what you don’t want.

Factorials

Factorials are crucial for calculating combinations, such as finding how many ways we can have 3 days with no rain out of 4 days.

A factorial (represented as \(n!\)) is the product of an integer and all the integers below it.

• For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\)

When using the binomial theorem, we calculate binomial coefficients using factorials.

For this problem, we used:

• \({4 \choose 3} = \frac{4!}{3!(4-3)!} = 4\)
• \({4 \choose 4} = \frac{4!}{4!(0!)},\) which simplifies to 1

Understanding factorials can greatly enhance the comprehension of probabilities and combinations.