Supposed (as is roughly correct) that each child born is equally likely to be a boy or a girl and that sexes of successive children are independent. If we let BG mean that the older child is a boy, and the younger child is a girl, then each of the combinations BB, BG, GB, GG has probability 0.25. Ashley and Brianna each have two children.
a.) You know that at least one of Ashley's children is a boy. What is the conditional probability that she has two boys?
b.) You know that Brianna's older child is a boy. What is the conditional probability that she has two boys?
please help, if you can :)
The formula of the conditional probability
P(X\Y)=P(XY)/P(Y)
a)Y="at least one is a boy"
Y=BB+BG+GB, P(Y)=3/4
X=BB
If X then Y => XY=X P(XY)=P(X)=1/4
P(X\Y)=(1/4)/(3/4)=1/3
b)Y="the older child is a boy"
Y=BB+BG, P(Y)=1/2
X=BB, XY=X, P(XY)=1/4
P(X/Y)=(1/4)/(1/2)=1/2
a.) Ah, the wonders of conditional probability! In this case, since we know that at least one of Ashley's children is a boy, we can eliminate the possibility of having two girls (GG). So, we are left with three possibilities: BB, BG, and GB. Each of these combinations has an equal probability of 0.25. However, since we want to find the probability that she has two boys, we only consider the BB combination.
Therefore, the probability that Ashley has two boys, given that she has at least one boy, is 1/3. (Because there is only one BB combination out of the three remaining possibilities.)
b.) Now, for Brianna's situation. We know that her older child is a boy, but we don't have any information about the younger child. So, we need to consider all the possibilities, except GG. We are left with BB, BG, and GB, each with a probability of 0.25.
Among these three possibilities, only BB satisfies the condition of having two boys. So, the conditional probability that Brianna has two boys, given that her older child is a boy, is 1/3.
In essence, both Ashley and Brianna have the same conditional probability of having two boys: 1/3. It's like a clown car full of mathematical possibilities!
a) Let's compute the conditional probability that Ashley has two boys given that at least one of her children is a boy.
Let A be the event that Ashley has two boys, and B be the event that at least one of Ashley's children is a boy.
We are interested in P(A|B), the conditional probability of A given B.
We can use Bayes' theorem to calculate this:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A) is the probability that Ashley has two boys, which is 0.25 (BB).
P(B|A) is the probability that at least one of Ashley's children is a boy given that she has two boys, which is 1 (since having two boys guarantees at least one boy).
P(B) is the probability that at least one of Ashley's children is a boy. This can be calculated as 1 - P(not B), where not B is the event that both of Ashley's children are girls (GG). Therefore, P(B) = 1 - P(GG) = 1 - 0.25 = 0.75.
Now we can calculate P(A|B):
P(A|B) = (1 * 0.25) / 0.75 = 0.25 / 0.75 = 1/3 = 0.333
Therefore, the conditional probability that Ashley has two boys given that at least one of her children is a boy is 0.333 or 1/3.
b) Let's compute the conditional probability that Brianna has two boys given that her older child is a boy.
Let B be the event that Brianna has two boys, and C be the event that Brianna's older child is a boy.
We are interested in P(B|C), the conditional probability of B given C.
Again, we can use Bayes' theorem to calculate this:
P(B|C) = (P(C|B) * P(B)) / P(C)
P(B) is the probability that Brianna has two boys, which is 0.25 (BB).
P(C|B) is the probability that Brianna's older child is a boy given that she has two boys, which is 1 (since having two boys guarantees that the older child is a boy).
P(C) is the probability that Brianna's older child is a boy. This is also 0.5 since the older child can be either a boy or a girl with equal probability.
Now we can calculate P(B|C):
P(B|C) = (1 * 0.25) / 0.5 = 0.25 / 0.5 = 0.5
Therefore, the conditional probability that Brianna has two boys given that her older child is a boy is 0.5 or 1/2.
Certainly! I can help you with these probability questions.
a.) To find the conditional probability that Ashley has two boys given that at least one of her children is a boy, we need to consider the possible combinations of children she can have. Since there are two children, there are four equally likely combinations: BB, BG, GB, and GG.
However, we know that at least one of Ashley's children is a boy. So we can eliminate the GG combination from our consideration. This means that there are three possible combinations left: BB, BG, and GB.
We are asked to find the probability of having two boys (BB) given that at least one child is a boy. This means we need to find the probability of having two boys (BB) out of the three remaining possible combinations (BB, BG, GB).
Since each combination is equally likely (probability 0.25), and there are three possible combinations, the probability of having two boys (BB) given that at least one child is a boy is 1/3 or approximately 0.333.
b.) To find the conditional probability that Brianna has two boys given that her older child is a boy, we need to consider the possible combinations of children she can have. Similar to the previous question, there are four equally likely combinations: BB, BG, GB, and GG.
However, we know that Brianna's older child is a boy. So we can eliminate the GB and GG combinations from our consideration. This means that there are two possible combinations left: BB and BG.
We are asked to find the probability of having two boys (BB) given that her older child is a boy. This means we need to find the probability of having two boys (BB) out of the two remaining possible combinations (BB, BG).
Since each combination is equally likely (probability 0.25), and there are two possible combinations, the probability of having two boys (BB) given that her older child is a boy is 1/2 or 0.5.
So, the conditional probability for Ashley is 1/3 or approximately 0.333, and the conditional probability for Brianna is 1/2 or 0.5.