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Assume that babies born are equally likely to be boys (B) or girls (G). Assume a woman has six children, none of whom are twins. Which sequence is more likely? Explain. Sequence A: GGGGGG Sequence B: GGGBBB

Short Answer

Expert verified
Both sequences A (GGGGGG) and B (GGGBBB) have the same probability of 0.015625.

Step by step solution

01

Understanding the problem

Given that babies born are equally likely to be boys (B) or girls (G), meaning the probability of having a girl child (P(G)) or a boy child (P(B)) is 0.5, we need to find the probability of each sequence (P(GGGGGG) and P(GGGBBB)). Each birth is an independent event.
02

Calculating the probability for Sequence A

The probability of having 6 girls (GGGGGG) can be calculated by P(GGGGGG) = P(G) x P(G) x P(G) x P(G) x P(G) x P(G), since each birth is an independent event. This means the total probability = \(0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.015625\).
03

Calculating the probability for Sequence B

The probability of having 3 girls and 3 boys (GGGBBB) is calculated by P(GGGBBB) = P(G) x P(G) x P(G) x P(B) x P(B) x P(B). As each birth is an independent event, we multiply the probabilities, giving \(0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.015625\).
04

Comparing the probabilities

After calculating, we find that both Sequence A and Sequence B have the same probability. This is due to the independence of each birth event and the equal likelihood of having a boy or a girl.

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Key Concepts

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

Independent Events
In genetics, independent events play a crucial role, especially when predicting the outcomes of traits in offspring. An independent event is one where the outcome does not influence or is not influenced by other events. For example, when flipping a coin, the result of one flip does not affect the result of the next flip.

In the context of our exercise, the probability of a child being a boy or a girl is independent of the gender of any previous children. This means that even if a woman has already had five boys, the probability that her next child will be a boy or a girl is still 0.5.

In genetics, this concept is essential when we consider Mendelian inheritance, where traits are passed from parents to offspring through independent alleles. Each allele pair segregates independently during the formation of gametes, leading to a variety of possible genetic outcomes.
Binomial Probability
The binomial probability is a type of probability calculation for a process where there are two possible outcomes (success or failure) in each of a series of independent trials. It follows a distribution known as the binomial distribution.

Using the principles behind the binomial distribution, we can calculate the probability of obtaining a specific number of successes (k) in a certain number of trials (n), given the probability of success in a single trial (p). The formula for binomial probability is:
\[ P(k; n, p) = \binom{n}{k}p^k(1-p)^{n-k} \]
where \(\binom{n}{k}\) represents the binomial coefficient.

In our exercise, 'success' could be defined as having a baby girl (G) with a probability of success (p) as 0.5. Therefore, the probability of any specific sequence of six children, such as having exactly three girls (GGGBBB), can be calculated using the binomial probability formula.
Probability Calculation
Probability calculation in genetics is used to determine the likelihood of various genetic outcomes. It is a mathematical way of representing the chances of a particular event occurring. The calculation typically considers the total number of possible outcomes and the number of ways a specific outcome can occur.

In our exercise, the probability of having either Sequence A (GGGGGG) or Sequence B (GGGBBB) involves calculating the likelihood of these sequences occurring under the assumption of independent events. Both sequences have the same number of both girls and boys, and with each gender having an equal probability of 0.5, the probability calculations yield the same result:
\[ P(GGGGGG) = P(G) \times P(G) \times P(G) \times P(G) \times P(G) \times P(G) \]\[ P(GGGBBB) = P(G) \times P(G) \times P(G) \times P(B) \times P(B) \times P(B) \]
After simplifying, both probabilities equal 0.015625. This simple calculation demonstrates that the sequence of genders in a family does not influence the likelihood of each possible sequence, as long as we assume equal probability and independent occurrences for each event.

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Most popular questions from this chapter

The sample space shows all possible sequences of child gender for a family with 3 children. The table is organized by the number of girls in the family. $$ \begin{array}{llll} \hline \text { 0 Girls } & \text { 1 Girl } & \text { 2 Girls } & \text { 3 Girls } \\ \hline \text { BBB } & \text { GBB } & \text { BGG } & \text { GGG } \\ \hline & \text { BGB } & \text { GBG } & \\ \hline & \text { BBG } & \text { GGB } & \end{array} $$ a. How many outcomes are in the sample space? b. If we assume all outcomes in the sample space are equally likely, find the probability of having the following numbers of girls in a family of 3 children: i. all 3 girls ii. no girls iii. exactly 2 girls

In California, about \(92 \%\) of teens who take the written driver's exam fail the first time they take it (www.teendrivingcourse.com). Suppose that Sam and Maria are randomly selected teenagers taking the test for the first time. a. What is the probability that they both pass the test? b. What is the probability that Sam OR Maria passes the test?

a. Explain how you could use a random number table (or the random numbers generated by software or a calculator) to simulate rolling a fair four-sided die 20 times. Assume you are interested in the probability of rolling a 1 . Then report a line or two of the random number table (or numbers generated by a computer or calculator) and the values that were obtained from it. b. Report the empirical probability of rolling a 1 on the four-sided die from part (a), and compare it with the theoretical probability of rolling a 1 .

A person was trying to figure out the probability of getting two heads when flipping two coins. He flipped two coins 10 times, and in 2 of these 10 times, both coins landed heads. On the basis of this outcome, he claims that the probability of two heads is \(2 / 10\), or \(20 \%\). Is this an example of an empirical probability or a theoretical probability? Explain.

Some estimates say that \(10 \%\) of the population is left-handed. We wish to design a simulation to find an empirical probability that if five babies are born on a single day, one or more will be left-handed. Suppose we decide that the even digits \((0,2,4,\), 6,8 ) will represent left-handed babies and the odd digits will represent right-handed babies. Explain what is wrong with the stated simulation method, and provide a correct method.

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