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Chapter 5: Problem 33

Taste Test for PTC The taste test for PTC (phenylthiocarbamide) is a favorite exercise for every human genetics class. It has been established that a single gene determines the characteristic, and that \(70 \%\) of Americans are "tasters," while \(30 \%\) are "nontasters." Suppose that 20 Americans are randomly chosen and are tested for PTC. a. What is the probability that 17 or more are "tasters"? b. What is the probability that 15 or fewer are "tasters"?

Short Answer

Expert verified
Question: Calculate the probability that (a) 17 or more Americans in a sample of 20 are "tasters" and (b) 15 or fewer Americans in a sample of 20 are "tasters", given that 70% of Americans are "tasters" and 30% are "nontasters". Solution: a. The probability that 17 or more are "tasters" can be calculated as the sum of binomial probabilities for 17, 18, 19, and 20 "tasters": \(P(\geq 17) = P(17) + P(18) + P(19) + P(20)\) b. The probability that 15 or fewer are "tasters" can be calculated as the sum of binomial probabilities from 0 to 15 "tasters": \(P(\leq 15) = P(0) + P(1) + \ldots + P(15)\)

Step by step solution

01

a. The probability that 17 or more are "tasters"

To find the probability of 17 or more "tasters" in a sample of 20 people, we can calculate it by finding the sum of the binomial probabilities for 17, 18, 19, and 20 "tasters". The binomial probability formula is given by: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\) Where \(n\) is the sample size (20 in this case), \(k\) is the number of successes (tasters), \(p\) is the probability of success (0.7), and \(\binom{n}{k}\) is the combination of choosing k successes from n trials. So, we need to calculate: \(P(17) + P(18) + P(19) + P(20)\)
02

Calculate the binomial probabilities for 17, 18, 19, and 20 tasters

Using the binomial probability formula: \(P(17) = \binom{20}{17} 0.7^{17} (1-0.7)^{20-17}\) \(P(18) = \binom{20}{18} 0.7^{18} (1-0.7)^{20-18}\) \(P(19) = \binom{20}{19} 0.7^{19} (1-0.7)^{20-19}\) \(P(20) = \binom{20}{20} 0.7^{20} (1-0.7)^{20-20}\)
03

Sum up the probabilities

Finally, sum up the probabilities: \(P(\geq 17) = P(17) + P(18) + P(19) + P(20)\)
04

b. The probability that 15 or fewer are "tasters"

To find the probability of 15 or fewer "tasters" in a sample of 20 people, we can calculate it by finding the sum of the binomial probabilities from 0 to 15 "tasters". So, we need to calculate: \(P(0) + P(1) + \ldots + P(15)\)
05

Calculate the binomial probabilities for 0 to 15 tasters

Using the binomial probability formula: \(P(k) = \binom{20}{k} 0.7^k (1-0.7)^{20-k}\) for \(k = 0, 1, \ldots, 15\)
06

Sum up the probabilities

Finally, sum up the probabilities: \(P(\leq 15) = P(0) + P(1) + \ldots + P(15)\)

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Key Concepts

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

Probability Distribution
Probability distribution represents all the possible values and likelihoods that a random variable can take within a given range. In the PTC taste test example, we're dealing with a discrete probability distribution because the number of 'tasters' among the 20 Americans can only be whole numbers from 0 to 20.

The distribution gives us a complete picture of the probability of each possible outcome. For instance, it allows us to understand not just the likelihood of a single event, like exactly 17 people being 'tasters', but also cumulative probabilities, such as 17 or more being 'tasters'.
Binomial Theorem
The binomial theorem provides a quick way to expand expressions that are raised to a power. In probability, the binomial theorem underpins the binomial distribution, which applies to scenarios with two possible outcomes: success ('taster') or failure ('non-taster'). Here, the relevant formula takes the general form
\(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)
\(n\) is the number of trials, \(k\) is the number of desired successes, \(p\) is the probability of a single success, and \(\binom{n}{k}\) is a binomial coefficient. The binomial theorem allows us to handle various probability calculations involving this type of scenario without expanding large binomial expressions.
Combination Mathematics
Combination mathematics, or 'combinatorics', involves the selection of items from a collection, such that (unlike permutations) the order of selection does not matter. In our example, we use combinations to calculate how many ways we can choose a certain number of 'tasters' from the total group of 20.

This is represented mathematically as \(\binom{n}{k}\), also read as 'n choose k'. Such calculations are integral to solving binomial probability problems because they measure the different ways an event can occur without regard to order.
Genetics in Probability
Genetics often relies on probability due to its predictive nature. Traits, such as being a 'taster' for PTC, can be modeled using binomial probability, as they typically have two outcomes governed by genetic predisposition. The exercise involving PTC taste testing illustrates how the principles of genetics can be analyzed using binomial probabilities. Since each person has an independent and constant probability of being a 'taster', the rules of binomial distribution apply, making the analysis of such genetic data possible and meaningful.

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Most popular questions from this chapter

Fast Food and Gas Stations Forty percent of all Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway. Suppose a random sample of \(n=25\) Americans who travel by car are asked how they determine where to stop for food and gas. Let \(x\) be the number in the sample who respond that they look for gas stations and food outlets that are close to or visible from the highway. a. What are the mean and variance of \(x ?\) b. Calculate the interval \(\mu \pm 2 \sigma\). What values of the binomial random variable \(x\) fall into this interval? c. Find \(P(6 \leq x \leq 14)\). How does this compare with the fraction in the interval \(\mu \pm 2 \sigma\) for any distribution? For mound-shaped distributions?

A packaging experiment is conducted by placing two different package designs for a breakfast food side by side on a supermarket shelf. The objective of the experiment is to see whether buyers indicate a preference for one of the two package designs. On a given day, 25 customers purchased a package from the supermarket. Let \(x\) equal the number of buyers who choose the second package design. a. If there is no preference for either of the two designs, what is the value of \(p,\) the probability that a buyer chooses the second package design? b. If there is no preference, use the results of part a to calculate the mean and standard deviation of \(x\). c. If 5 of the 25 customers choose the first package design and 20 choose the second design, what do you conclude about the customers' preference for the second package design?

Evidence shows that the probability that a driver will be involved in a serious automobile accident during a given year is .01. A particular corporation employs 100 full-time traveling sales reps. Based on this evidence, use the Poisson approximation to the binomial distribution to find the probability that exactly two of the sales reps will be involved in a serious automobile accident during the coming year.

Bacteria in Water Samples If a drop of water is placed on a slide and examined under a microscope, the number \(x\) of a particular type of bacteria present has been found to have a Poisson probability distribution. Suppose the maximum permissible count per water specimen for this type of bacteria is five. If the mean count for your water supply is two and you test a single specimen, is it likely that the count will exceed the maximum permissible count? Explain.

Let \(x\) be a Poisson random variable with mean \(\mu=2 .\) Calculate these probabilities: a. \(P(x=0)\) b. \(P(x=1)\) c. \(P(x>1)\) d. \(P(x=5)\)
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