Question

If two carriers of the gene for albinism marry, each of their children has probability \(\frac{1}{4}\) of being albino (see Example 3.6.1). If such a couple has six children, what is the probability that (a) none will be albino? (b) at least one will be albino? [Hint: Use part (a) to answer part (b); note that "at least one" means "one or more."]

Short answer

The probability that none of the six children will be albino is approximately 0.178. The probability that at least one child will be albino is approximately 0.822.

Step-by-step solution

Calculate the probability of none being albino

The probability of one child being albino is given as \( \frac{1}{4} \). Therefore, the probability of one child not being albino is \( 1 - \frac{1}{4} = \frac{3}{4} \). Since each child is independent of the others, the probability of six children not being albino is \( \left(\frac{3}{4}\right)^6 \).

Compute the result for part (a)

To find the probability that none of the six children will be albino, raise the single event probability of not being albino to the power of 6. This is \( \left(\frac{3}{4}\right)^6 \approx 0.178 \).

Understand the concept of 'at least one'

'At least one' is the complement of 'none'. Therefore, to compute the probability that at least one child will be albino, subtract the probability of none from 1.

Calculate the probability of at least one being albino

Using the complement rule, the probability that at least one child will be albino is \( 1 - \left(\frac{3}{4}\right)^6 \approx 1 - 0.178 \approx 0.822 \).

Key concepts

Genetic Inheritance

Genetic inheritance is the process by which genetic information is passed on from parents to their offspring. Each parent contributes one allele for every gene, resulting in a pair of alleles for each gene in the offspring. In the case of albinism, it is a recessive genetic trait, meaning that a child needs to inherit the albino allele from both parents to express the trait. When parents are carriers, they have one albino allele and one normal allele. The probability that a child inherits the albino allele from one parent is 1 in 2, or \( \frac{1}{2} \). Since the child must inherit the allele from both parents to be albino, the probability of a child being albino is \( \frac{1}{2} \) times \( \frac{1}{2} \) equals \( \frac{1}{4} \).

This simple Mendelian concept can be understood as flipping two coins, your aim is for both coins to land on tails. Each coin (similar to each allele from a parent) independently has the same probability of landing on heads or tails. If tails represents the recessive albino allele, then getting two tails – one from each coin – is akin to a child being albino.

Binomial Probability

Binomial probability refers to the probability of obtaining a fixed number of successes in a fixed number of independent trials, with the same probability of success on each trial. Considering the textbook problem, each childbirth is an independent trial, and producing an albino child is considered 'success' in genetic terms. With the probability of success being \( \frac{1}{4} \) for each child, we can use the binomial probability formula to calculate the chances of different outcomes.

For instance, the probability of exactly three albino children out of six can be calculated using the binomial probability formula\( P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \), where:\

• \( n \) is the total number of trials (children),
• \( k \) is the number of successful trials (albino children),
• \( p \) is the probability of success on a single trial,
• \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials.

Examples of binomial distribution:

When calculating the likelihood that a specific number of children will be albino, you are constructing a binomial distribution. For instance, the probability of exactly one child being albino out of six is reflected in a distinct point on this distribution.

Complement Rule

The complement rule is a fundamental principle in probability theory which states that the probability of an event not happening is equal to one minus the probability of the event happening. Put simply, if you're curious about the chances of at least one occurrence of an event, it's often easier to calculate the probability of the event not occurring at all and then subtracting that from one.

This approach is especially useful when the 'at least one' occurrences include multiple scenarios or are difficult to calculate individually. For the albinism example, it is easier to find the probability of no albino children and subtract it from one to find the probability of at least one albino child. The calculation for the textbook problem using the complement rule is a neat example, where \(1 - \left(\frac{3}{4}\right)^6 \) gives us the probability of having at least one albino child among six. The complement rule is particularly helpful in situations with a large number of trials or when assessing risk in genetic counseling.