A box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate. I am sharing the box with three of my friends. The four of us take turns, each one drawing a chocolate at random from those available in the box. Yes, we're drawing without replacement. We intend to eat the chocolates, not put them back.
Politely, I let my friends draw before me.
For each of the subparts, provide your answer as an unsimplified fractional expression.

a) What is the chance that I draw a dark chocolate?
b) Given that none of my friends draws a dark chocolate, what is the chance that I draw a dark chocolate?
c) What is the chance that all four of us draw milk chocolates?
d) What is the chance that the friend who gets to select first draws a dark chocolate and I draw a dark chocolate too?

Step-by-step explanation:

Given that a box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate.

a)  the chance that I draw a dark chocolate

b) Given that none of my friends draws a dark chocolate,  the chance that I draw a dark chocolate is

since now all 5 dark chocolates would be there and total changed to 9 because 3 friends took 1 each.

c) All four drawn milk chocolates would be

d) the chance that the friend who gets to select first draws a dark chocolate and I draw a dark chocolate too

=P(first friend draws a dark chocolate) * P(I draw a dark chocolate)

When it comes to me chances are either there are 4 dark, or 3 dark, or 1 dark depending upon the other two friends drew black or not

=

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

c

c

c

c

c

c

c

cc

c

step-by-step explanation: