1. Answer (fully simplified):
–16384x²¹ – 57344x¹⁸y⁴ – 86016x¹⁵y⁸ – 71680x¹²y¹² – 35840x⁹y¹⁶ – 10752x⁶y²⁰ – 1792x³y²⁴ – 128y²⁸
Alternate answer with everything simplified except the binomial coefficients:
–16384x²¹ – ₇C₁(8192x¹⁸y⁴) – ₇C₂(4096x¹⁵y⁸) – ₇C₃(2048x¹²y¹²) – ₇C₄(1024x⁹y¹⁶) – ₇C₅(512x⁶y²⁰) – ₇C₆(256x³y²⁴) – 128y²⁸
Please note that you would not earn full credit if you left the "powers of a power" unexpanded.

2. Approx. –1.305 ´ 10¹⁸ a²p⁹³

3. Recall Pascal’s Triangle (the top row, with the single 1 in it, corresponds to n = 0):

In the n = 3 row, which is marked with an arrow above, the coefficients are . You can see that they add up to 8, which is what the equation claimed (since 2³ = 8). Thus the equation is true when n = 3.

In the n = 7 row, which is the row with the 7 in it, you can see that 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128, which equals 2⁷. That proves that the equation is also true when n = 7.

To prove the equation in general, you have to apply a clever trick of writing the RHS of the equation as 2ⁿ = (1 + 1)ⁿ. Now perform a binomial expansion to get , which simplifies to , which is the LHS of the equation as desired.

4.(a) 1/13 (no work required)
(b) 1/17 (work required)
(c) 352/833 (work required)

5.(a) ₇₂C₄ = 1,028,790
(b) ₃₉P₆ = 2,349,088,560
(c) 39⁶ = 3,518,743,761

6.(a) 0.06 (work required)
If we randomly select a person at a random time, there is a 6% chance that what we choose is a rainy day AND a person who is using an umbrella. [Note: The multiplication rule is valid here since the umbrella probability is actually P(B | A), the probability of B assuming that A is true.]

(b) 0.85 (work required)
If we randomly select a time, there is an 85% chance that rain is not falling.

(c) 0.6 (work required)
If we randomly select a person (assuming that rain is falling), there is a 60% chance that the person is umbrella-less.

(d) 0.09 (work required)
If we randomly select a person at a random time, there is a 9% chance that what we choose is a rainy day AND a person who is NOT using an umbrella. [Note: The multiplication rule is valid here since the umbrella probability is actually P(~B | A), the probability of ~B assuming that A is true.]

7.(a) 0.0009 (work required)
(b) 0.9409 (work required)
(c) 0.0591 (work required)

8.(a) Approx. 0.04755 (work required)
(b) Approx. 0.95245 (work required)

9. One rather complicated example would be the event of rolling a fair die and obtaining a value that is a divisor of 6. You can surely come up with something simpler. Justification:
P(divisor of 6) = P(1 or 2 or 3 or 6) = P(Œ È  È Ž È ‘) = P(Œ) + P() + P(Ž) + P(‘) = 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3. The addition rule is valid because the four successful die outcomes are mutually exclusive.