Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, "sickle" shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that:
(a) two will have the disease?
(b) none will have the disease?
(c) at least one will neither have the disease nor be a carrier?
(d) the first child with the disease will the be 3rd child?
To answer these probability questions, we need to use the principles of probability and basic calculations. Let's break down each question and calculate the probabilities step by step:
(a) Two children will have the disease:
The probability of a child having the disease is 25% or 0.25, as stated in the problem. So, the probability of two children having the disease is calculated by multiplying the individual probabilities:
P(two children have the disease) = P(child 1 has the disease) * P(child 2 has the disease)
P(two children have the disease) = 0.25 * 0.25 = 0.0625 or 6.25%
(b) None of the children will have the disease:
The probability of a child having neither the disease nor being a carrier is also stated in the problem, which is 25% or 0.25. To calculate the probability that none of the children will have the disease, we can use the complement rule:
P(none have the disease) = 1 - P(at least one has the disease)
Since we haven't calculated the probability of at least one child having the disease yet, we'll calculate it later in question (c).
(c) At least one child will neither have the disease nor be a carrier:
This question is essentially the complement of question (b). To calculate the probability of at least one child being disease-free and not a carrier, we can use the complement rule as stated previously:
P(at least one neither has the disease nor is a carrier) = 1 - P(none have the disease)
P(at least one neither has the disease nor is a carrier) = 1 - 0.25 = 0.75 or 75%
(d) The first child with the disease will be the third child:
Since the probability of a child having the disease is independent of the order of birth, the probability of the first child with the disease being the third child is simply the probability of any child having the disease, which is 25% or 0.25.
To summarize:
(a) Two children will have the disease: 6.25%
(b) None of the children will have the disease: 25%
(c) At least one child will neither have the disease nor be a carrier: 75%
(d) The first child with the disease will be the third child: 25%
To calculate the probabilities, we can use the rules of probability. Let's break down each question step by step:
(a) What is the probability that two children will have the disease?
If both parents are carriers of the disease, then each parent has a 50% chance of passing down the sickle cell gene to their child.
Let's calculate the probability of two children having the disease:
Probability of a child having the disease = 25% = 0.25
Since the events (having the disease or not having the disease) are independent for each child, we can multiply the probabilities:
Probability of two children having the disease = Probability of one child having the disease multiplied by itself = 0.25 * 0.25 = 0.0625
Therefore, the probability that two children will have the disease is 0.0625 or 6.25%.
(b) What is the probability that none of the children will have the disease?
Since the probability of a child having the disease is 25% or 0.25, the probability of a child not having the disease is 100% - 25% = 75% = 0.75.
Probability of none of the children having the disease = Probability of each child not having the disease multiplied by itself = 0.75 * 0.75 * 0.75 = 0.421875
Therefore, the probability that none of the children will have the disease is 0.421875 or 42.1875%.
(c) What is the probability that at least one child will neither have the disease nor be a carrier?
The probability that at least one child will neither have the disease nor be a carrier is equal to 1 minus the probability that all three children will either have the disease or be carriers.
Probability of all three children having the disease or being carriers = Probability of having the disease * Probability of being a carrier = 0.25 * 0.5 = 0.125
Therefore, the probability that at least one child will neither have the disease nor be a carrier = 1 - 0.125 = 0.875
Therefore, the probability that at least one child will neither have the disease nor be a carrier is 0.875 or 87.5%.
(d) What is the probability that the first child with the disease will be the third child?
Since each child is independent of the others, the probability of any specific child having the disease remains 0.25 or 25% for each child.
Therefore, the probability that the first child with the disease will be the third child is the same as the probability that the first child has the disease and the second child does not.
Probability of the first child having the disease = 0.25
Probability of the second child not having the disease = 1 - 0.25 = 0.75
Probability that the first child with the disease will be the third child = Probability of the first child having the disease multiplied by the probability of the second child not having the disease = 0.25 * 0.75 = 0.1875
Therefore, the probability that the first child with the disease will be the third child is 0.1875 or 18.75%.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
a. .25 * .25 * .75 = ?
b. .75^3 = ?
I'll leave c and d for you.