Q3. A bag contains 7 red balls and 3 green balls. A ball is taken out replaced. A second ball is then taken out.
i.
Find the following probabilities?
a. P(R)
b. P(G)
c. P (G and G)
d. P (R and R)
e. P (R and G)
f. P (G and R)

Part (a)

There are 7 red out of 7+3 = 10 total

Part (b)

We have 3 green out of 10 total

Part (c)

3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)

So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.

Part (d)

Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.

(7/10)*(7/10) = 49/100

Part (e)

7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.

(7/10)*(3/10) = 21/100

Part (f)

We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph

step-by-step explanation:

infinity is the answer