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Chapter 4: Problem 27

Find the probability and answer the questionsWhen Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the expected value of \(3 / 4,\) as Mendel claimed?

Short Answer

Expert verified
The probability of a green pea is approximately 0.7379, which is reasonably close to Mendel's expected probability of 0.75.

Step by step solution

01

- Determine Total Sample Size

Add the number of green peas and yellow peas to find the total sample size. The total sample size is the sum of green peas (428) and yellow peas (152). So, 428 + 152 = 580.
02

- Calculate the Probability of Green Peas

To find the probability of a pea being green, divide the number of green peas by the total sample size. Using the numbers, the probability is \( \frac{428}{580} \).
03

- Simplify the Probability

Simplify the fraction \( \frac{428}{580} \) by dividing both the numerator and the denominator by their greatest common divisor (which is 4). Thus, \( \frac{428 \div 4}{580 \div 4} = \frac{107}{145} \).
04

- Convert Fraction to Decimal

Convert the fraction \( \frac{107}{145} \) to a decimal by performing the division. \( \frac{107}{145} \approx 0.7379 \). This means the probability of getting a green pea is approximately 0.7379.
05

- Compare with Expected Probability

The expected probability according to Mendel is \( \frac{3}{4} \) or 0.75. Compare 0.7379 with 0.75. Since they are very close, we can say that the observed probability is reasonably close to the expected probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mendel's genetics
Mendel's genetics is the study of how traits are inherited from one generation to the next. Gregor Mendel, through his work on pea plants, discovered the fundamental laws of inheritance. He found that traits are determined by pairs of genes, one from each parent, and these traits can be either dominant or recessive. This foundational work is critical for understanding genetic probability in biology.
probability calculation
Probability calculation helps us determine how likely an event is to occur. In Mendel's experiments, we calculate the probability of getting a green pea by dividing the number of green peas by the total number of peas. Probability ranges from 0 to 1, where 0 means the event will not happen, and 1 means it will definitely happen. For example, the probability of selecting a green pea from 428 green and 152 yellow peas is calculated as follows: \(\frac{428}{580}\).
sample size analysis
Sample size analysis is crucial in statistics because a larger sample size can give a more accurate estimate of the true probability. In Mendel's experiment, our sample consists of 580 peas, combining 428 green and 152 yellow peas. Larger samples tend to yield results that are closer to what is expected based on theoretical probabilities, minimizing the effect of random variations.
fraction simplification
Fraction simplification makes probabilities easier to understand and compare. In our example, the probability of getting a green pea is initially \( \frac{428}{580} \). To simplify, we find the greatest common divisor (GCD) of 428 and 580, which is 4. Dividing both the numerator and the denominator by 4 gives us \( \frac{107}{145} \).
decimal conversion
Decimal conversion translates a fraction into its decimal form, making it easier to interpret and compare. To convert \( \frac{107}{145} \) into a decimal, we perform the division \( 107 \div 145 \approx 0.7379 \). This indicates that the probability of getting a green pea is approximately 0.7379, which we can compare to the expected probability of 0.75. The close values show that our results align well with Mendel's expected probability.

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Most popular questions from this chapter

When the horse California Chrome won the 140th Kentucky Derby, a \(\$ 2 dollars bet on a California Chrome win resulted in a winning ticket worth \)\7 dollars. a. How much net profit was made from a \(\$ 2\) win bet on California Chrome? b. What were the payoff odds against a California Chrome win? c. Based on preliminary wagering before the race, bettors collectively believed that California Chrome had a 0.228 probability of winning. Assuming that 0.228 was the true probability of a Califonia Chrome victory, what were the actual odds against his winning? d. If the payoff odds were the actual odds found in part (c), what would be the worth of a \(\$ 2\) win ticket after the California Chrome win?

Express all probabilities as fractions. In a horse race, a quinela bet is won if you selected the two horses that finish first and second, and they can be selected in any order. The 140 th running of the Kentucky Derby had a field of 19 horses. What is the probability of winning a quinela bet if random horse selections are made?

Express all probabilities as fractions. If radio station call letters must begin with either \(\mathrm{K}\) or \(\mathrm{W}\) and must include either two or three additional letters, how many different possibilities are there?

Find the probability.At Least One. In Exercises \(5-12,\) find the probability. It has been reported that \(20 \%\) of iPhones manufactured by Foxconn for a product launch did not meet Apple's quality standards. An engineer needs at least one defective iPhone so she can try to identify the problem(s). If she randomly selects 15 iPhones from a very large batch, what is the probability that she will get at least 1 that is defective? Is that probability high enough so that she can be reasonably sure of getting a defect for her work?

Probability of At Least One Let \(A=\) the event of getting at least 1 defective iPhone when 3 iPhones are randomly selected with replacement from a batch. If \(5 \%\) of the iPhones in a batch are defective and the other \(95 \%\) are all good, which of the following are correct? a. \(P(\bar{A})=(0.95)(0.95)(0.95)=0.857\) b. \(P(A)=1-(0.95)(0.95)(0.95)=0.143\) c. \(P(A)=(0.05)(0.05)(0.05)=0.000125\)
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