It should be b let me know if im right
Here, you have to combine like terms. So, it is 11m-3v.
There are no equations? Well there’s no picture
Let's start from what we know.

Note that:

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first

with only positive trems (squares of even numbers) and second

with negative (squares of odd numbers). So:

And now the proof.
1) n is even.
In this case, both

and

have

terms. For example if n=8 then:

Generally, there will be:

Now, calculate our sum:



So in this case we prove, that:

2) n is odd.
Here,

has more terms than

. For example if n=7 then:

So there is

terms in

,

terms in

and:

Now, we can calculate our sum:




We consider all possible n so we prove that:
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
P(brown) = 12% = 0.12
P(Yellow) = 15% = 0.15
P(Red) = 12% = 0.12
P(blue) = 23% = 0.23
P(orange) = 23% = 0.23
P(green) = 15% = 0.15
A.) Compute the probability that a randomly selected peanut M&M is not yellow.
P(not yellow) = P(Yellow)' = 1 - P(Yellow) = 1 - 0.15 = 0.85
B.) Compute the probability that a randomly selected peanut M&M is brown or red.
P(Brown) or P(Red) :
0.12 + 0.12 = 0.24
C.) Compute the probability that three randomly selected peanut M&M’s are all brown.
P(brown) * P(brown) * P(brown)
0.12 * 0.12 * 0.12 =0.001728
D.) If you randomly select three peanut M&M’s, compute that probability that none of them are blue.
P(3 blue)' = 1 - P(3 blue)
P(3 blue) = 0.23 * 0.23 * 0.23 = 0.012167
1 - P(3 blue) = 1 - 0.012167 = 0.987833
If you randomly select three peanut M&M’s, compute that probability that at least one of them is blue.
P(1 blue) + p(2 blue) + p(3 blue)
(0.23) + (0.23*0.23) + (0.23*0.23*0.23)
0.23 + 0.0529 + 0.012167
= 0.295067