5

Find the probability that, in a group of 5000 newborn white babies in the United States, at least 4 babies suffer from cystic fibrosis....

Question

Find the probability that, in a group of 5000 newborn white babies in the United States, at least 4 babies suffer from cystic fibrosis.

Find the probability that, in a group of 5000 newborn white babies in the United States, at least 4 babies suffer from cystic fibrosis.



Answers

Assume
that
the
probability
of the
birth
of
a
child
of
a
particular
sex
is $ 50\% $.
In
a
family
with
four
children,
what
are
the
probabilities
that(a)
all
the
children
are
boys,
(b)
all
the
children
are
the same
sex,
and
(c)
there
is
at
least
one
boy?

So we're looking at Okay. Family of Eline feature Pam, liberal leaning Trump Charles are trying to figure out with their feet, but have cystic fibrosis or not And they're both of the parents have history of cystic fibrosis and their families. So let's start with the lean we know about Elaine Could be a carrier since her brother had cystic fibrosis. But because her parents didn't have Mr Quite boats is their chant types are gonna be hetero Zika's. So there came here. This is a fibrosis is a process of disease, So people with genotype two small ants would have cystic fibrosis. Okay, so we're crossing her parents, but had her brother have to stick my roses and away the lane might how they're excessive gene or might not have the gene. So this is something that we need to determine, and she was trying to start a family. Charles Charles is definitely carrier because when he had a baby with his first wife, also was a carrier, their child waas suffering from the disease. And the main question that we need to ask answer is whether their baby would have cystic fibrosis or not. So the probability of Elaine. Being a curier is to third. So this is Elaine Charles. The probability off. Um, alien meaning carrier. So the rain IHS, hear hear is 2/3. How did we get that? So she doesn't have does not have cystic fibrosis. Right? So no see, Eliminates one out of four possibilities for her. Right? Um, four minus one. No, no. Three possibilities. And out of three. So three possibilities. There are only two chances of being carrier. Because there's to Gina types. You have. And for and then 34 possibilities possibilities. And she would have two chances. So this is where the 2/3 came from. And the probability off, baby having the cystic fibrosis disease. So, baby, uh, cystic fibrosis. If we consider that Elaine has the, um, cystic level sister, she's a carrier than the probability of a baby having the disease. It's 14 because we were crossing and end with big and small and and the probability of the baby being have existed for a process with who she is too small ends. It's 1/4. This is where quarter comes in. The total probability will be two birds times 14 and I will give you do 12. Which ISS once it

This problem says that a married man and a woman both happened gene for cystic fibrosis. And so, according to some laws of genetics, the probability that their first child will develop cystic fibrosis is 0.25 So in other words, there's a 25% chance that if they have a child, that child will have the disease. Um, so is asking to explain what this probability means, which is exactly that. Uh, if they have a child, the chance the child has the disease is 0.2 is 25%. There's a 25% chance, Um, as a long run relative probability. Another way of thinking about this would be that if they have many Children, if they have many Children, about 1/4 would have the disease. Now why? But why is it about 1/4 and not exactly 1/4? Well, that's what Part B is asking moms. And this is saying, why does it not mean that if the couple has four Children, one of them is guaranteed to get cystic fibrosis? Well, because, uh, one important a probability does not mean exactly 1/4 of the Children will have the disease. And in fact, we can calculate the probability that if, uh all four Children do not have the disease, well, this is equal to a 0.75 to the fourth power because each child has 75% chance of not having the disease. And if we assume that each of them are independent, the probability that each one has the disease is independent. Then this is just point someone five to the power for since there's four Children, so we see there's a non zero probability that all none of them have the disease.

82. According to a recent article, the average number of babies born with significant hearing loss is approximately two per 1000 babies and healthy baby nursery. The number climbs to an average of 30 for 1000 babies in intensive care. Nursery. Suppose that 1000 babies from healthy baby nurseries were randomly selected. Find the probability that exactly two babies were born death. In this setting, we're dealing with the binomial distribution. We can consider either a baby being selected to be deaf or not death. And we have a set number of trials. So we're gonna say that X there's gonna be binomial with 1000 trials and the probability of a success or the probability of being deaf in this case is too out of 1000. The question is ultimately asking for what is the probability that exactly two babies were born death. Now, in this case, we can use our T A calculator and the distributions drop down menu. Find our binomial PDS function in the binomial pdf function. It'll tell us a single value for any of the situations of this binomial in which we have 1000 trials with the probability of two over 1000 and we're looking for exactly two to be death so he could play this and a calculator will find that the probability is 0.27

Okay, This question asks us probability of having at least five boys from a group of five Children assuming equal chances for each gender. So to find the probability of at least three, well, that can happen if we have three boys or if we have floor or if we have five, and we can just add these probabilities together because they're mutually exclusive. So now we'll just use the binomial distribution to find each of these, although we don't have to do it for T of five, because there's only one way that can happen, so we can just use a short cut on that one. But for P of three, it's five Children choosing three to be boys times the probability to get a boy, and we want three of them. And then that means that the other two Children are girls. And then next one, we have five Children picking four to be boys times the chance to be a boy to the fourth power times the remaining child, which is a girl. And then we add the probability of getting five boys, which is just that 1/2 chance happening five times in a row, so Now let's simplify. So we have five Choose three, which using our calculator for that five shoes. Three is 10. So this is tend about by 32 then five. Choose Force five suite of 5/32 and then we have a 1/32 so again were adding P three plus before, plus P five. And that's 16/32 which is 1/2 and that's our probability.


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