Each of two parents has the genotype red divided by blond, which consists of the pair of alleles that determine hair color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one red allele, that color will dominate and the child's hair color will be red. a. List the different possible outcomes. Assume that these outcomes are equally likely. b. What is the probability that a child of these parents will have the blond divided by blond genotype? c. What is the probability that the child will have red hair color? a. List the possible outcomes. A. red divided by red comma red divided by blond comma blond divided by red comma and blond divided by blond B. red divided by red comma red divided by blond comma and blond divided by blond C. red divided by blond and blond divided by red D. red divided by red and blond divided by blond b. The probability that a child of these parents will have the blond divided by blond genotype is nothing. (Round to two decimal places as needed.) c. The probability that the child will have red hair color is nothing. (Round to two decimal places as needed.)

Step-by-step explanation:

a. The possible outcome

From the given information, the parents have the genotype red/blond and the pair of allele determines the hair color.

Let R represents red

Let B represent blond

Check attachment for crossing.

So, we have four outcomes which are red/red, red/blond, blond/red, blond/blond.

correct option is A

red divided by red comma red divided by blond comma blond divided by red comma and blond divided by blond

b. Probability of having the same blond is given as

Here, there are 4 outcomes red/red, red/blond, blond/red, blond/blond

Therefore,

The blond/blond occurs once out of the the four possible outcomes, then the probability of the parent having blond blond is

P(blond/blond)=1/4

P(blond/blond)= ¼

c. Probability of red hair colour is

P(red/anything)

Here, there are 4 outcomes red/red, red/blond, blond/red, blond/blond

P(red/anything)=3/4=0.75

red occurs three time and it is only once that red did not occur out of 4 total outcome.

P(red/anything) =3/4

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!