Question

it is not uncommon for the husband and wife in a married couple to carry the gene for cystic fibrosis without having the disease themselves. suppose we select one of these couples at random. according to the laws of genetics, the probability that any child they have will develop cystic fibrosis is 0.25.
(a) explain what this probability means.
o if you select a couple who has four children, it is guaranteed that one of their children will develop cystic fibrosis.
o if you select a very large random sample of couples like these, about 25% of their children will develop cystic fibrosis.
o if you select a small random sample of couples like these, about 2.5 of their children will develop cystic fibrosis.
o if you select 100 random couples like these, exactly 25 of the couples will have one child who will develop cystic fibrosis.
(b) if researchers randomly select 4 such couples, is one of these couples guaranteed to have a first child who develops cystic fibrosis? explain your answer.
this probability describes what happens in repetitions of a chance process.

Explanation:

Step1: Understanding probability concept

Probability of \(0.25\) (or \(25\%\)) for a single - couple's child having cystic fibrosis is a long - term relative frequency. It doesn't guarantee specific outcomes in small samples.

Step2: Analyzing sample - size impact

• For a very large sample:

By the law of large numbers, as the sample size \(n\) increases, the sample proportion \(\hat{p}\) approaches the population proportion \(p\). If \(p = 0.25\), in a very large random sample of couples, about \(25\%\) of the children will develop cystic fibrosis.

• For a small sample (e.g., \(n = 4\)):

The binomial probability formula is \(P(X=k)={n\choose k}p^{k}(1 - p)^{n - k}\), where \(n = 4\), \(p=0.25\), and \(k\) is the number of successes (children with cystic fibrosis). The probability that exactly \(k = 1\) child (out of \(4\)) has cystic fibrosis is \({4\choose1}(0.25)^{1}(0.75)^{3}=\frac{4!}{1!(4 - 1)!}\times0.25\times0.421875=4\times0.25\times0.421875 = 0.421875\), and there are other possibilities (\(k = 0\), \(k = 2\), \(k=3\), \(k = 4\)). So, we can't guarantee that exactly \(1\) (or \(25\%\) of \(4\)) child will have cystic fibrosis.

• For \(n = 100\):

Using the normal approximation to the binomial (since \(np=100\times0.25 = 25\) and \(n(1 - p)=100\times0.75 = 75\) are both \(\geq5\)), the mean of the binomial distribution is \(\mu=np = 25\) and the standard deviation is \(\sigma=\sqrt{np(1 - p)}=\sqrt{100\times0.25\times0.75}=\sqrt{18.75}\approx4.33\). While the expected number of children with cystic fibrosis is \(25\), there is still a distribution around this mean (e.g., using the empirical rule, about \(95\%\) of the values lie in the interval \(\mu\pm2\sigma=(25 - 2\times4.33,25 + 2\times4.33)=(16.34,33.66)\)), so it's not guaranteed that exactly \(25\) children will have cystic fibrosis.

• For a single couple with four children:

Each child's outcome (having cystic fibrosis or not) is an independent event. The probability that a child has cystic fibrosis is \(p = 0.25\), but we can't guarantee that one of the four children will have it. For example, the probability that none of the four children have cystic fibrosis is \((0.75)^{4}=0.31640625\).

Answer:

(a) A probability of \(0.25\) means that in the long - run (over a large number of independent repetitions, such as considering a very large number of couples having children), about \(25\%\) of the children will develop cystic fibrosis. It is a measure of the likelihood of an event (a child developing cystic fibrosis) occurring in a single trial (a couple having a child), but it does not guarantee a specific outcome in a small - scale or single - couple situation.
(b) No. When researchers randomly select \(4\) couples, since each couple's children's cystic - fibrosis status is an independent binomial event (with \(n = 4\) trials per couple and \(p=0.25\) probability of a child having cystic fibrosis per trial), using the binomial probability formula \(P(X = k)={n\choose k}p^{k}(1 - p)^{n - k}\) (where \(n = 4\), \(p = 0.25\)), there are multiple possible values of \(k\) (number of children with cystic fibrosis in a family of \(4\) children). For example, \(P(X = 0)=(0.75)^{4}=0.31640625\), \(P(X=1)={4\choose1}(0.25)^{1}(0.75)^{3}=0.421875\), etc. So, we cannot be guaranteed that one of the \(4\) couples will have a first child who develops cystic fibrosis.