# What is the probability that I will get a disease

An individual’s cancer risk has a lot to do with other factors, such as his or her age. For instance, a woman’s lifetime risk of developing colon and rectal cancer is just over 5 percent, or about 523 cases per 10,000 women. Thanks for using ChaCha. [ Source: http://www.chacha.com/question/what-is-the-probability-that-i-will-get-a-disease ]

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**What is the probability that I will get a disease**“- What is the probability that I will get a disease
- http://www.chacha.com/question/what-is-the-probability-that-i-will-get-a-disease
- An individual’s cancer risk has a lot to do with other factors, such as his or her age. For instance, a woman’s lifetime risk of developing colon and rectal cancer is just over 5 percent, or about 523 cases per 10,000 women. Thanks for usin…

- What is the probability of the offspring inheriting sickle cell d…?
- http://wiki.answers.com/Q/What+are+the+chances+someone+with+sickle+cell+can+pass+to+their+offspring
- Everyone carries genes that are responsible for creating hemoglobin. Some people carry the defective sickle cell hemoglobin gene. If you and your partner both have this sickle cell gene, it will be passed on to your child. As a result, your…

- Is probability for acquiring metastatic bone disease affected by …?
- http://www.orthop.washington.edu/uw/metastatic/tabID__3368/ItemID__291/PageID__3/qview__true/Articles/Default.aspx
- Certain behaviors, life factors, and habits can put people at greater risk for the kinds of cancers that might lead to metastatic bone disease, but not at greater risk for metastatic bone disease itself. For example, smokers are more likely…

## Related Questions Answered on Y!Answers

- Can anyone help me with this probability problem?
- Q: Suppose that the probability that a child produced by a couple will have a particular disease is 1/10. If they plan to have four children, what is the probability that one or more children will the disease?The answer is .3439 but I can’t figure out how to get that. The formulas we’ve covered for this section are P(A and B), P(A or B), and 1-P(A) for the complement.
- A: Use the complement rule:First, determine the probability that no children have the disease.P(no disease) = (.9)^4 = .6561P(at least one diseased child) = 1 – .6561 = .3439Good luck in your studies,~ Mitch ~

- statistics homework dealing with probability?
- Q: I can not figure out how they are getting this answer. If anyone can help I would be very greatful!Question: Suppose that probabilitiy that a child produced by a couple will have a particular disease is 1/10. If they plan to have four children, what isthe probability that one or more children will have the disease?Answer: .3439Thanks!
- A: Let X be the number of kids withe the disease. X has the binomial distribution with n = 4 trials and success probability p = 0.1 In general, if X has the binomial distribution with n trials and a success probability of p thenP[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)for values of x = 0, 1, 2, …, nP[X = x] = 0 for any other value of x.The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n – x failures.Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.X ~ Binomial( n = 4 , p = 0.1 )the mean of the binomial distribution is n * p = 0.4 the variance of the binomial distribution is n * p * (1 – p) = 0.36 the standard deviation is the square root of the variance = √ ( n * p * (1 – p)) = 0.6 The Probability Mass Function, PMF, f(X) = P(X = x) is:P( X = 0 ) = 0.6561 P( X = 1 ) = 0.2916 P( X = 2 ) = 0.0486 P( X = 3 ) = 0.0036 P( X = 4 ) = 1e-04 P( X ≥ 1 ) = 1 – P(X = 0) = 1 – 0.6561 = 0.3439 P(at least one) is the same as saying 1 – P(zero)

- Probability question in genetics?
- Q: Galactosemia is a recessive human disease that is treatable by restricting lactose and glucose in the diet. Susan and her husband are both heterozygous for the galactosemia gene.a)Susan is pregnant with twin. If she has fraternal (nonidentical) twins, what is the probability obth of the twins will be girls who have galact. ?b) If the twins are identical, what is the probability that both will be girls and have galact.?For parts c and d, assume that none of the children is a twin.c) if Susan and her husband have four children, what is the probability that none of the four will have galact?d) If the couple has four children, what is the probability that atleast one child will have the disease?Note: I really dont get the twin stuff. I mean, how does it differ when its with twins?? Can you plz help me with the problem???
- A: r&s mom gave a nice answer for the genetics.The difference between fraternal and identical twins is that in fraternal twins there are two eggs fertilized (so each is a genetically separate individual) and the two may have different probabilities of having a particular trait. With identical twins, the egg was fertilized, then during split apart during one of the cell divisions, so the two resulting individuals are genetically identical (not necessarily completely, but don’t worry about those complicating factors), so genetically, they will have the same chance of having the same trait.So, in a), if Susan and her husband are both Gg, then the probability that they have two children with galactosemia is the probability of each child multiplied together: 1/4 x 1/4 = 1/16However, the problem also states that the kids are both girls (1/2 x 1/2), so, you have to put in that probability also:(probability of galactosemia x probability of a girl x probability of galactosemia x probability of a girl):1/4 x 1/2 x 1/4 x 1/2 = 1/64b) If the twins are identical, then they can be treated as a single individual genetically, so the probability of have identical twins that are both girls and have galactosemia is:1/4 x 1/2 = 1/8For c and d, assume none of the children is a twin.c) The probability that one child does not have galactosemia is 3/4 (GG, Gg, or Gg). So, the probability that none of four children will have galactosemia is 3/4 x 3/4 x 3/4 x 3/4 = 81/256c) The probability that at least one child has the disease out of four total is the probability of gg mulitplied by the probability of G- (GG, Gg, or Gg) for the other three children:1/4 x 3/4 x 3/4 x 3/4 = 27/256