Two litters of a particular rodent species have been born, one with two brownhaired and one gray-haired (litter 1), and the other with three brown-haired and two gray-haired (litter 2). We select a litter at random and then select an offspring at random from the selected litter.

(a) What is the probability that the animal chosen is brown-haired?
(b) Given that a brown-haired offspring was selected, what is the probability that the sampling was from litter 1?

a. So probability that the animal chosen is brown-haired = 0.633

b. Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A) = 0.5263.

Step-by-step explanation:

Being that We are given,event of Brown hair with two disjoint events, one is { ( BrownHair ) ∩ ( Litter 1) } and the other is { ( BrownHair ) ∩ ( Litter 2) } .

a) To find the probability that the animal chosen is brown-haired,

Let A : we choose a brown-haired rodent and B : we choose litter1.

So using the axioms of probability, we can write

P(A) = P(A | B) * P(B) + P(A | Bc) * P(Bc)

Making use of the given information, we get;

number of brown haired rodents in litter 2 P(AB) Total number of rodents in litterl

and

P(A |B^{c}) =\frac{\text{number of brown haired rodents in litter2}}{\text{Total number of rodents in litter2}}= \frac{3}{5}

And also it is given that we choose litter at random ,so P(B) = P(Bc ) = 1/2

So now we plug these values in the equation of P(A) and get

P(A) = (\frac{2}{3}*\frac{1}{2}) + (\frac{3}{5}*\frac{1}{2}) = \frac{2}{6}+\frac{3}{10} = 0.633

So probability that the animal chosen is brown-haired = 0.633

b) Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A)

Lets make use of Bayes rule to find this conditional probability,

So using theorem we get,

P(B|A) = \frac{P(A|B)*P(B)}{P(A|B)*P(B)+P(A|B^{c})*P(B^{c})}

P(B|A) = \frac{(1/2)*(2/3)}{[(1/2)*(2/3)]+[(1/2)*(3/5)]} = \frac{10}{19} = 0.5263

Thus, Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A) = 0.5263.

part a

in a city ice cream is mostly only bought when the temperature is very high as ice cream cools down our

so we can derive the conclusion that as the temperature increases the number of ice creams sold also increases

part b

y intercept is 5

slope is 2

happy to

pls mark as brainliest

chicken chicken chicken

step-by-step explanation: