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Find the probability.At Least One. In Exercises \(5-12,\) find the probability. In the United States, the true probability of a baby being a boy is 0.512 (based on the data available at this writing). Among the next six randomly selected births in the United States, what is the probability that at least one of them is a girl?

Short Answer

Expert verified
0.9791

Step by step solution

01

Identify the given values

The probability of a baby being a boy, denoted as P(Boy), is 0.512. Therefore, the probability of a baby being a girl, P(Girl), is 1 - P(Boy) = 1 - 0.512 = 0.488. The number of births, n, is 6.
02

Define 'at least one' in probability terms

'At least one' can be found using the complement rule. Calculate the probability that there is no girl (all boys), and then subtract this from 1 to find the probability that there is at least one girl.
03

Calculate the probability of having no girls

The probability of all six babies being boys is found by raising the probability of one boy to the power of 6. Therefore, P(All Boys) = P(Boy)^6 = 0.512^6.
04

Compute the numerical value

Calculate 0.512^6. Using a calculator, 0.512^6 鈮 0.0209.
05

Calculate the probability of at least one girl

The probability of at least one girl is then given by P(At least one girl) = 1 - P(All Boys) = 1 - 0.0209.
06

Final calculation

Subtract to find the final probability: 1 - 0.0209 = 0.9791.

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Key Concepts

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

Complement Rule
In probability, the complement rule is a handy tool. It helps us find the likelihood of an event happening by looking at the opposite case. In our problem, we need to find the probability of having at least one girl among six babies. Instead of directly calculating this, we first calculate the chance of having no girls (only boys) and then subtract this from 1. This approach is effective because it's often easier to calculate the probability of the complement. Specifically, the complement rule is stated as: \( P(A') = 1 - P(A) \), where \( P(A') \) is the probability of the event not occurring. Here, our event is having at least one girl, and the complement of this event is having no girls at all.
Binomial Probability
The binomial probability formula is crucial when we have a fixed number of trials with two possible outcomes: success and failure. In our example, each birth is a trial with two outcomes: boy or girl. We're interested in at least one success (girl) in 6 trials (births).
The base formula for binomial probability is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \], where \( p \) is the success probability (girl being 0.488), \( n \) is the total number of trials (6), and \( k \) is the number of successes. Although calculating directly with binomial might be tedious, using the complement rule with binomial probability for zero successes (all boys) simplifies our task.
Statistical Analysis
Statistical analysis involves using data to make predictions and decisions. It helps us figure out the likelihood of outcomes, like having at least one girl in six births.
First, we identified the key probabilities: 0.512 for a boy and 0.488 for a girl. Then, we chose the complement rule to simplify our calculations. By raising the probability of one boy to the power of 6, we found the chance of having no girls (0.0209). Subtracting this from 1 showed the probability of having at least one girl (0.9791).
These steps show how statistical methods break complex problems into manageable tasks.
Probability of Events
Understanding probability involves breaking down the chances of different outcomes. In our example, we examine the probability of having various combinations of boys and girls in six births. By using the complement rule and binomial probability, we efficiently solve the problem.
Probabilities are the backbone of statistical analysis, helping us predict scenarios and make informed decisions. We learned that calculating the probability of complementary events is often simpler. This means looking at what we don't want, and understanding that the universe of outcomes must add up to 1.

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Most popular questions from this chapter

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?

Express all probabilities as fractions. In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning 鈥渂lackjack鈥 hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. The two cards can be in any order.Find the probability of being dealt a blackjack hand. What approximate percentage of hands arewinning blackjack hands?

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. \(P(A)=(0.05)(0.05)(0.05)=0.000125\)

Express all probabilities as fractions. Quicken Loans offered a prize of \(\$ 1\) billion to anyone who could correctly predict the winner of the NCAA basketball tournament. After the "play-in" games, there are 64 teams in the tournament. a. How many games are required to get 1 championship team from the field of 64 teams? b. If you make random guesses for each game of the tournament, find the probability of picking the winner in every game.

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?

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