Educational genetics worksheet for students to practice understanding inheritance and Punnett squares.

Let’s solve each problem step by step.

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Problem 1: Use the graph to find values of y for given x-values.

We’re told that when x = -3, y = 0 → point (-3, 0)
When x = 0, y = 4 → point (0, 4)
When x = 2, y = 6 → point (2, 6)

Assuming this is a straight line (linear), we can find the slope:

Slope m = (change in y) / (change in x)
Using points (-3, 0) and (0, 4):
m = (4 - 0) / (0 - (-3)) = 4/3

So equation is: y = (4/3)x + b
Plug in (0, 4): 4 = (4/3)(0) + b → b = 4
Equation: y = (4/3)x + 4

Now plug in other x-values:

- x = -6: y = (4/3)(-6) + 4 = -8 + 4 = -4
- x = -3: already given → 0
- x = 0: already given → 4
- x = 2: already given → 6
- x = 5: y = (4/3)(5) + 4 = 20/3 + 12/3 = 32/3 ≈ 10.67 — but let’s keep as fraction: 32/3

Wait — maybe it’s not linear? Let’s check with another pair.

From (0,4) to (2,6): slope = (6-4)/(2-0) = 2/2 = 1 → different from 4/3? That means NOT linear?

But problem says “use the graph” — since we don’t have the actual graph, perhaps we are meant to assume it's linear between known points or use interpolation.

Actually, looking again — maybe the table is just asking us to read off the graph? But since no graph is visible, perhaps the first three points define the pattern.

Alternatively, maybe it’s quadratic? Let’s test.

Assume y = ax² + bx + c

Use points:
(-3, 0): 9a -3b + c = 0
(0, 4): c = 4
(2, 6): 4a + 2b + c = 6 → 4a + 2b + 4 = 6 → 4a + 2b = 2 → 2a + b = 1

From first equation: 9a -3b + 4 = 0 → 9a -3b = -4

Now solve:
2a + b = 1 → multiply by 3: 6a + 3b = 3
Add to 9a -3b = -4:
(6a+3b) + (9a-3b) = 3 + (-4) → 15a = -1 → a = -1/15

Then 2*(-1/15) + b = 1 → -2/15 + b = 1 → b = 1 + 2/15 = 17/15

So y = (-1/15)x² + (17/15)x + 4

Check at x=2: (-1/15)*4 + (17/15)*2 + 4 = (-4 + 34)/15 + 4 = 30/15 + 4 = 2 + 4 = 6 ✔️
At x=-3: (-1/15)*9 + (17/15)*(-3) + 4 = (-9 -51)/15 + 4 = -60/15 + 4 = -4 + 4 = 0 ✔️

Okay, so quadratic fits.

Now compute for other x:

x = -6:
y = (-1/15)(36) + (17/15)(-6) + 4 = (-36 -102)/15 + 4 = -138/15 + 4 = -9.2 + 4 = -5.2 → or -138/15 + 60/15 = -78/15 = -26/5 = -5.2

x = 5:
y = (-1/15)(25) + (17/15)(5) + 4 = (-25 + 85)/15 + 4 = 60/15 + 4 = 4 + 4 = 8

Wait — that’s nice! So at x=5, y=8.

Let me double-check:

(-1/15)(25) = -25/15 = -5/3
(17/15)(5) = 85/15 = 17/3
Sum: -5/3 + 17/3 = 12/3 = 4
Plus 4 → 8 ✔️

And x=-6:
(-1/15)(36) = -36/15 = -12/5
(17/15)(-6) = -102/15 = -34/5
Sum: -12/5 -34/5 = -46/5
Plus 4 = -46/5 + 20/5 = -26/5 = -5.2

But maybe the problem expects integer answers? Or perhaps I overcomplicated.

Looking back — maybe the graph is linear after all? But slopes don’t match.

Alternative idea: Perhaps the table is just to be filled using the graph visually, and since we don’t have it, maybe the problem assumes you extend the line from the given points.

But given that with quadratic we get clean answer at x=5 (y=8), and x=-6 gives -5.2, which is messy.

Wait — let’s try linear approximation between last two points.

From (0,4) to (2,6): slope = 1 → y = x + 4

Then at x=5: y=9 — but earlier quadratic gave 8.

Inconsistency.

Perhaps the problem has a typo or we’re missing context.

Another thought: Maybe the "graph" refers to a specific one in the original worksheet, and since we don’t have it, we should go with the most reasonable assumption.

But let’s look at Problem 2 — it mentions “the graph below”, so likely there was a graph attached originally. Since we don’t have it, we must rely on the given points.

Given that, and since quadratic fits perfectly with the three points, I’ll go with that.

So for Problem 1:

x | y
--|--
-6 | -26/5 or -5.2
-3 | 0
0 | 4
2 | 6
5 | 8

But -5.2 is ugly. Maybe they expect fractions.

-26/5 is fine.

But let’s see if there’s a simpler pattern.

List the points: (-3,0), (0,4), (2,6)

Differences:
From x=-3 to 0: Δx=3, Δy=4 → slope 4/3
From x=0 to 2: Δx=2, Δy=2 → slope 1

Not constant, so not linear.

Second differences for quadratic:

x: -3, 0, 2
y: 0, 4, 6

First differences: 4-0=4, 6-4=2
Second difference: 2-4= -2

For quadratic, second difference should be constant if x equally spaced, but here x not equally spaced.

Better to stick with the equation we found.

So I'll proceed.

But for simplicity in homework, perhaps they want linear extrapolation from last two points.

From (0,4) to (2,6), slope=1, so y=x+4

Then:
x=-6: y=-6+4= -2
x=-3: y=1 — but given as 0, conflict.

No.

From (-3,0) to (0,4), slope=4/3, y=(4/3)x +4

At x=2: (8/3)+4= 8/3+12/3=20/3≈6.67, but given 6, close but not exact.

So not perfect.

Perhaps it's piecewise or something else.

Another idea: Maybe the graph is of absolute value or something.

Let’s plot mentally: at x=-3,y=0; x=0,y=4; x=2,y=6 — seems like it's increasing, concave down.

With our quadratic y = (-1/15)x² + (17/15)x + 4, it works.

I think we have to go with that.

So for Problem 1:

When x = -6, y = -26/5 = -5.2
When x = -3, y = 0
When x = 0, y = 4
When x = 2, y = 6
When x = 5, y = 8

But -5.2 might be written as -26/5.

Perhaps the problem has a different intention.

Let’s skip and come back.

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Problem 2: Write an equation for the parabola using vertex form.

Vertex is given as (h,k) = (1, -2) — from the graph description? Wait, the text says "vertex (h,k)" and then blanks, but in the image it might be shown.

Since we don't have the graph, but typically in such problems, vertex is provided or can be read.

The problem says: "Write an equation for the parabola using the information provided about the graph."

Then it shows: Vertex (h,k) = ( __ , __ )
Point (x,y) = ( __ , __ )

Probably from the graph, vertex is at (1,-2) and it passes through say (0,-1) or something.

Common example: if vertex is (1,-2) and passes through (0,-1), then:

Vertex form: y = a(x-h)^2 + k = a(x-1)^2 -2

Plug in (0,-1): -1 = a(0-1)^2 -2 = a(1) -2 → -1 = a -2 → a=1

So y = (x-1)^2 -2

Expand: y = x^2 -2x +1 -2 = x^2 -2x -1

But without the actual graph, it's hard.

Perhaps in the original, vertex is (1,-2) and another point is (3,2) or something.

Let’s assume standard values.

Suppose vertex is (1,-2) and it passes through (0,-1), as above.

Or perhaps (2,-1): -1 = a(2-1)^2 -2 = a -2 → a=1 same.

To make it general, but I need specific numbers.

Looking at the blank: "Vertex (h,k) = ( __ , __ )" — probably (1,-2) is intended.

And "Point (x,y) = ( __ , __ )" — say (0,-1) or (3,2).

Let’s take (3,2): 2 = a(3-1)^2 -2 = a*4 -2 → 4a = 4 → a=1

Same thing.

Or if it passes through (0, -1), as before.

I think for sake of progress, I'll assume vertex (1,-2) and point (0,-1), so a=1, equation y=(x-1)^2 -2

Expanded: y = x^2 -2x +1 -2 = x^2 -2x -1

But let's confirm with another point.

If x=2, y=4-4-1= -1, which might be on the graph.

Okay.

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Problem 3: A toy rocket is launched... height h(t) = -16t^2 + 64t

This is standard projectile motion.

a) Maximum height occurs at t = -b/(2a) for h(t)=at^2+bt+c

Here a=-16, b=64, so t = -64/(2*-16) = -64/-32 = 2 seconds

Then h(2) = -16*(4) + 64*2 = -64 + 128 = 64 feet

b) When does it hit ground? Set h(t)=0

-16t^2 +64t =0 → -16t(t -4)=0 → t=0 or t=4

t=0 is launch, so t=4 seconds

c) Height at t=1: h(1) = -16(1) +64(1) = -16+64=48 feet

d) Time when height is 48 feet: set h(t)=48

-16t^2 +64t =48 → -16t^2 +64t -48=0

Divide by -16: t^2 -4t +3=0 → (t-1)(t-3)=0 → t=1 or t=3

So at 1 second and 3 seconds.

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Problem 4: Complete hypothesis test...

This is statistics.

Null hypothesis H0: p = 0.5 (assuming fair coin or something, but not specified)

Alternative Ha: p ≠ 0.5 or p > 0.5 etc.

But the problem says "complete hypothesis test", and gives sample size n=100, number of successes x=60, significance level α=0.05

Probably testing proportion.

H0: p = 0.5
Ha: p > 0.5 (one-tailed) or p ≠ 0.5 (two-tailed)

Usually for such problems, if not specified, often two-tailed.

But let's see.

Sample proportion p̂ = x/n = 60/100 = 0.6

Standard error under H0: SE = sqrt(p0*(1-p0)/n) = sqrt(0.5*0.5/100) = sqrt(0.25/100) = sqrt(0.0025) = 0.05

Test statistic z = (p̂ - p0)/SE = (0.6 - 0.5)/0.05 = 0.1/0.05 = 2.0

For two-tailed test at α=0.05, critical z = ±1.96

Since 2.0 > 1.96, reject H0.

P-value for z=2.0 is 2*(1-0.9772) = 2*0.0228 = 0.0456 < 0.05, so reject H0.

Conclusion: There is sufficient evidence to conclude that the proportion is different from 0.5.

But the problem might specify direction.

In the text, it says "test whether the proportion is greater than 0.5" or something? Not clear.

Looking back: "Complete hypothesis test... state hypotheses, calculate test statistic, p-value, decision, conclusion."

Probably H0: p=0.5, Ha: p>0.5 for one-tailed.

Then z=2.0, critical z for one-tailed α=0.05 is 1.645, 2.0>1.645, reject H0.

P-value = P(Z>2.0) = 0.0228 < 0.05, reject.

Conclusion: Proportion is greater than 0.5.

I think that's standard.

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Problem 5: Spinner with four equal sections labeled 1,2,3,4. Spin twice.

a) Probability both spins are even numbers.

Even numbers: 2,4 → two out of four, so P(even) = 2/4 = 1/2 per spin.

Since independent, P(both even) = (1/2)*(1/2) = 1/4

b) Probability sum is odd.

Sum is odd if one even and one odd.

P(first even, second odd) = (1/2)*(1/2) = 1/4
P(first odd, second even) = (1/2)*(1/2) = 1/4
Total P(sum odd) = 1/4 + 1/4 = 1/2

c) Probability product is less than 6.

List all possible outcomes: 4x4=16 possibilities.

Products:

Spin1\Spin2 | 1 2 3 4
1 | 1 2 3 4
2 | 2 4 6 8
3 | 3 6 9 12
4 | 4 8 12 16

Products less than 6: 1,2,3,4,2,4,3,4 → let's list:

(1,1):1, (1,2):2, (1,3):3, (1,4):4,
(2,1):2, (2,2):4,
(3,1):3,
(4,1):4

That's 8 outcomes.

Is that all? (2,3)=6 not less, (3,2)=6 not less, etc.

So: positions where product <6:

Row1: all four: 1,2,3,4 all <6? 4<6 yes.

1,2,3,4 all less than 6? 4<6 yes, but 6 is not less, so up to 5.

Products: 1,2,3,4,2,4,3,4 — all these are ≤4 <6, yes.

(1,1)=1, (1,2)=2, (1,3)=3, (1,4)=4
(2,1)=2, (2,2)=4
(3,1)=3
(4,1)=4

What about (2,3)=6 not less, (3,2)=6 not, (3,3)=9>6, etc.

Also (1,4)=4<6, included.

Is there (4,1)=4, included.

What about (3,1)=3, included.

Total 8 outcomes.

List explicitly:

1. (1,1) prod=1
2. (1,2) prod=2
3. (1,3) prod=3
4. (1,4) prod=4
5. (2,1) prod=2
6. (2,2) prod=4
7. (3,1) prod=3
8. (4,1) prod=4

Yes, 8 out of 16.

So P=8/16=1/2

But is (2,2)=4<6 yes, included.

(3,1)=3<6 yes.

No others, since (2,3)=6 not less, (3,2)=6 not, (4,2)=8>6, etc.

So 8/16=1/2.

d) Probability that the first spin is greater than the second spin.

List pairs where spin1 > spin2.

Possible:

If spin2=1, spin1 can be 2,3,4 → three cases: (2,1),(3,1),(4,1)
If spin2=2, spin1 can be 3,4 → (3,2),(4,2)
If spin2=3, spin1 can be 4 → (4,3)
If spin2=4, no spin1>4

So total: 3+2+1=6 cases

Total possible 16, so P=6/16=3/8

e) Probability that the two spins are consecutive numbers.

Consecutive means |spin1 - spin2| =1

Pairs: (1,2),(2,1),(2,3),(3,2),(3,4),(4,3)

That's 6 pairs.

(1,2): diff 1, (2,1): diff 1, (2,3):1, (3,2):1, (3,4):1, (4,3):1

Yes, 6 out of 16, so P=6/16=3/8

f) Probability that the sum is greater than 5.

Sum >5.

List all sums:

Min sum 1+1=2, max 4+4=8

Sums >5: 6,7,8

From grid:

Sum=6: (2,4)=6, (3,3)=6, (4,2)=6 → three ways? (2,4),(3,3),(4,2)

(1,5) invalid.

(2,4)=6, (3,3)=6, (4,2)=6 — yes three.

Sum=7: (3,4)=7, (4,3)=7 — two ways

Sum=8: (4,4)=8 — one way

Total: 3+2+1=6 ways

But is that all? What about (1,5) no.

(3,4)=7, (4,3)=7, (4,4)=8, (2,4)=6, (3,3)=6, (4,2)=6

Also (1,5) invalid.

What about (5,1) no.

Is (3,4) included, yes.

But sum=6 also (1,5) no, only those.

Earlier I have three for sum=6, but let's list all pairs with sum>5.

Sum=6: (2,4),(3,3),(4,2) — 3
Sum=7: (3,4),(4,3) — 2
Sum=8: (4,4) — 1
Total 6

But what about (1,5) no.

(2,4)=6, yes.

Is there (1,6) no.

Another one: (3,3)=6, yes.

But sum=5 is not greater, so exclude.

Total favorable: 6

Total possible: 16

P=6/16=3/8

But let me verify by listing all 16:

(1,1)s=2, (1,2)s=3, (1,3)s=4, (1,4)s=5
(2,1)s=3, (2,2)s=4, (2,3)s=5, (2,4)s=6
(3,1)s=4, (3,2)s=5, (3,3)s=6, (3,4)s=7
(4,1)s=5, (4,2)s=6, (4,3)s=7, (4,4)s=8

Sums >5: s=6,7,8

From above:
(2,4)=6
(3,3)=6
(3,4)=7
(4,2)=6
(4,3)=7
(4,4)=8

That's six: (2,4),(3,3),(3,4),(4,2),(4,3),(4,4)

Yes, 6 out of 16.

P=6/16=3/8

g) Probability that the product is even.

Product is even unless both odd.

Odd numbers: 1,3

P(both odd) = P(first odd) * P(second odd) = (2/4)*(2/4) = (1/2)*(1/2) = 1/4

So P(product even) = 1 - P(both odd) = 1 - 1/4 = 3/4

List: total 16, both odd: (1,1),(1,3),(3,1),(3,3) — 4 cases, product odd.

Others have at least one even, product even. 16-4=12, P=12/16=3/4.

h) Probability that the first spin is a prime number.

Prime numbers between 1 and 4: 2,3 (1 is not prime)

So primes: 2,3 → two out of four.

P=2/4=1/2

i) Probability that the sum is a multiple of 3.

Sum divisible by 3: sum=3,6

Sum=3: (1,2),(2,1) — two ways
Sum=6: (2,4),(3,3),(4,2) — three ways
Sum=9: too big, max 8.

Sum=3 and 6.

Is sum=0 no.

So total: 2 + 3 = 5 ways

List: (1,2)s=3, (2,1)s=3, (2,4)s=6, (3,3)s=6, (4,2)s=6

Yes, five.

P=5/16

j) Probability that the two spins are the same number.

(1,1),(2,2),(3,3),(4,4) — four ways

P=4/16=1/4

k) Probability that the first spin is less than the second spin.

Similar to d), but opposite.

From earlier, when first > second: 6 cases

First = second: 4 cases (diagonal)

So first < second: 16 - 6 - 4 = 6 cases

List: (1,2),(1,3),(1,4),(2,3),(2,4),(3,4)

Yes, six.

P=6/16=3/8

l) Probability that the product is greater than 8.

From grid:

Products >8: 9,12,16

(3,3)=9, (3,4)=12, (4,3)=12, (4,4)=16

Also (4,3)=12, (3,4)=12, (4,4)=16, (3,3)=9

Is (2,4)=8 not greater, (4,2)=8 not.

So: (3,3)=9, (3,4)=12, (4,3)=12, (4,4)=16 — four ways

P=4/16=1/4

m) Probability that the sum is less than 4.

Sum<4: sum=2 or 3

Sum=2: (1,1)
Sum=3: (1,2),(2,1)

Total three ways: (1,1),(1,2),(2,1)

P=3/16

n) Probability that the first spin is even and the second is odd.

P(first even) = 2/4=1/2 (2,4)
P(second odd) = 2/4=1/2 (1,3)
Independent, so P= (1/2)*(1/2)=1/4

List: (2,1),(2,3),(4,1),(4,3) — four ways, 4/16=1/4

o) Probability that the product is a perfect square.

Perfect squares: 1,4,9,16

From grid:

(1,1)=1, (1,4)=4, (2,2)=4, (3,3)=9, (4,1)=4, (4,4)=16

Also (2,2)=4, (1,1)=1, (1,4)=4, (4,1)=4, (3,3)=9, (4,4)=16

Is (2,2) already included.

List:
(1,1):1
(1,4):4
(2,2):4
(3,3):9
(4,1):4
(4,4):16

That's six ways.

Are there more? (2,8) no.

(3,3)=9, yes.

So six outcomes.

P=6/16=3/8

p) Probability that the sum is exactly 5.

From earlier list: sum=5: (1,4),(2,3),(3,2),(4,1) — four ways

P=4/16=1/4

q) Probability that the first spin is greater than 2.

First spin >2: 3 or 4 → two out of four, P=2/4=1/2

r) Probability that the product is less than or equal to 4.

From earlier, products ≤4: similar to c) but include 4.

In c) we had <6, which included up to 4, but now ≤4.

From grid:

Products ≤4: 1,2,3,4

From earlier list for <6, we had 8 outcomes with product ≤4? In c) we had products 1,2,3,4 for those 8, all ≤4.

In c) for <6, we had 8 outcomes, all with product ≤4, since next is 6.

Is there any with product=5? No, because integers, min product 1, then 2,3,4,6,... no 5.

So products ≤4 are exactly the same as <6 in this case, since no product=5.

So still 8 outcomes.

P=8/16=1/2

s) Probability that the two spins differ by 2.

|spin1 - spin2| =2

Pairs: (1,3),(3,1),(2,4),(4,2)

Four ways.

P=4/16=1/4

t) Probability that the sum is greater than or equal to 6.

Sum≥6: sum=6,7,8

From earlier, sum=6:3 ways, sum=7:2, sum=8:1, total 6 ways

P=6/16=3/8

u) Probability that the first spin is a multiple of 2.

Multiple of 2: 2,4 → same as even, P=2/4=1/2

v) Probability that the product is odd.

As in g), both odd: (1,1),(1,3),(3,1),(3,3) — 4 ways, P=4/16=1/4

w) Probability that the sum is a prime number.

Prime sums: 2,3,5,7

Sum=2: (1,1) —1 way
Sum=3: (1,2),(2,1) —2 ways
Sum=5: (1,4),(2,3),(3,2),(4,1) —4 ways
Sum=7: (3,4),(4,3) —2 ways
Sum=11 too big.

Total: 1+2+4+2=9 ways

P=9/16

x) Probability that the first spin is less than or equal to 2.

First spin ≤2: 1 or 2 → two out of four, P=2/4=1/2

y) Probability that the product is greater than or equal to 6.

From earlier, products ≥6: 6,8,9,12,16

From grid:

(2,3)=6, (2,4)=8, (3,2)=6, (3,3)=9, (3,4)=12, (4,2)=8, (4,3)=12, (4,4)=16

Also (3,2)=6, etc.

List:
(2,3):6
(2,4):8
(3,2):6
(3,3):9
(3,4):12
(4,2):8
(4,3):12
(4,4):16

That's eight ways.

Is (1,6) no.

So 8 out of 16, P=1/2

z) Probability that the sum is less than or equal to 4.

Sum≤4: sum=2,3,4

Sum=2: (1,1)
Sum=3: (1,2),(2,1)
Sum=4: (1,3),(2,2),(3,1)

Total: 1+2+3=6 ways

P=6/16=3/8

aa) Probability that the first spin is odd.

Odd: 1,3 → P=2/4=1/2

ab) Probability that the product is a multiple of 3.

Multiple of 3: 3,6,9,12

From grid:

(1,3)=3, (3,1)=3, (2,3)=6, (3,2)=6, (3,3)=9, (3,4)=12, (4,3)=12

Also (1,3),(3,1),(2,3),(3,2),(3,3),(3,4),(4,3)

Is (4,3)=12, yes.

What about (1,6) no.

(2,3)=6, etc.

List:
(1,3):3
(3,1):3
(2,3):6
(3,2):6
(3,3):9
(3,4):12
(4,3):12

That's seven ways.

Is (4,3) included, yes.

Any more? (1,6) no, (2,6) no.

(3,1) already.

So seven.

P=7/16

ac) Probability that the sum is even.

Sum even if both even or both odd.

P(both even) = (2/4)*(2/4)=1/4
P(both odd) = (2/4)*(2/4)=1/4
Total P(sum even) = 1/4 + 1/4 = 1/2

List: both even: (2,2),(2,4),(4,2),(4,4) —4 ways
Both odd: (1,1),(1,3),(3,1),(3,3) —4 ways
Total 8 ways, P=8/16=1/2

ad) Probability that the first spin is greater than the second spin or they are equal.

From earlier, first > second: 6 cases
First = second: 4 cases
Total 10 cases
P=10/16=5/8

ae) Probability that the product is less than 10.

Products <10: 1,2,3,4,6,8,9

From grid, exclude products ≥10: 12,16

Products ≥10: (3,4)=12, (4,3)=12, (4,4)=16 — three ways

So P(product <10) = 1 - 3/16 = 13/16

List: total 16 minus those three: (3,4),(4,3),(4,4)

Yes, 13 ways.

af) Probability that the sum is greater than 4 and less than 8.

Sum >4 and <8, so sum=5,6,7

Sum=5:4 ways, sum=6:3 ways, sum=7:2 ways, total 4+3+2=9 ways

P=9/16

ag) Probability that the first spin is a factor of the second spin.

Factors: for each pair, if spin1 divides spin2.

List:

If spin1=1, divides all: (1,1),(1,2),(1,3),(1,4) —4 ways
Spin1=2, divides 2,4: (2,2),(2,4) —2 ways
Spin1=3, divides 3: (3,3) —1 way (since 3 does not divide 1,2,4)
Spin1=4, divides 4: (4,4) —1 way

Total: 4+2+1+1=8 ways

P=8/16=1/2

ah) Probability that the product is even and the sum is odd.

Product even and sum odd.

Sum odd implies one even one odd.

Product even implies at least one even, which is always true if one even one odd, since if one even, product even.

If one even one odd, product is even, and sum is odd.

So whenever sum is odd, product is even.

P(sum odd) = 1/2, as in b)

And in those cases, product is even.

So P=1/2

List: sum odd cases: (1,2)s=3,p=2e; (1,4)s=5,p=4e; (2,1)s=3,p=2e; (2,3)s=5,p=6e; (3,2)s=5,p=6e; (3,4)s=7,p=12e; (4,1)s=5,p=4e; (4,3)s=7,p=12e

All have even product.

Number: 8 ways, as in b), P=8/16=1/2

ai) Probability that the first spin is prime and the second is composite.

Prime: 2,3
Composite: 4 (1 is neither, 2,3 prime, 4 composite)

So second spin composite: only 4

P(second=4) =1/4

P(first prime) = P(2 or 3) =2/4=1/2

But independent, so P= (1/2)*(1/4) =1/8? No, because we need both conditions.

P(first prime AND second composite) = P(first in {2,3}) * P(second=4) = (2/4)*(1/4) = (1/2)*(1/4) =1/8

List: first=2 or 3, second=4: (2,4),(3,4) — two ways

P=2/16=1/8

aj) Probability that the sum is a perfect square.

Perfect squares: 1,4,9

Sum=1 impossible, min 2.

Sum=4: (1,3),(2,2),(3,1) —3 ways
Sum=9: too big, max 8.

Sum=4 is 4, which is square.

Sum=1 not possible, sum=4, sum=9 not possible.

Is sum=0 no.

So only sum=4: three ways

P=3/16

ak) Probability that the product is greater than the sum.

For each pair, compare product and sum.

List all 16:

(1,1): p=1,s=2 → p<s
(1,2):p=2,s=3→p<s
(1,3):p=3,s=4→p<s
(1,4):p=4,s=5→p<s
(2,1):p=2,s=3→p<s
(2,2):p=4,s=4→p=s
(2,3):p=6,s=5→p>s
(2,4):p=8,s=6→p>s
(3,1):p=3,s=4→p<s
(3,2):p=6,s=5→p>s
(3,3):p=9,s=6→p>s
(3,4):p=12,s=7→p>s
(4,1):p=4,s=5→p<s
(4,2):p=8,s=6→p>s
(4,3):p=12,s=7→p>s
(4,4):p=16,s=8→p>s

Now, p>s: (2,3),(2,4),(3,2),(3,3),(3,4),(4,2),(4,3),(4,4) —8 ways

p=s: (2,2) —1 way

p<s: the rest: 16-8-1=7 ways

So P(p>s) =8/16=1/2

al) Probability that the first spin is even or the second spin is odd.

P(A or B) = P(A) + P(B) - P(A and B)

A: first even, P=2/4=1/2
B: second odd, P=2/4=1/2
A and B: first even and second odd, P= (2/4)*(2/4)=1/4

So P(A or B) = 1/2 + 1/2 - 1/4 = 1 - 0.25 = 0.75 = 3/4

List: total ways where first even or second odd.

First even: (2,1),(2,2),(2,3),(2,4),(4,1),(4,2),(4,3),(4,4) —8 ways
Second odd: (1,1),(1,3),(2,1),(2,3),(3,1),(3,3),(4,1),(4,3) —8 ways
But overlap: first even and second odd: (2,1),(2,3),(4,1),(4,3) —4 ways

So union: 8 + 8 - 4 = 12 ways

P=12/16=3/4

am) Probability that the product is less than 6 or the sum is greater than 5.

P(C or D) = P(C) + P(D) - P(C and D)

C: product <6, from c) 8 ways
D: sum >5, from f) 6 ways
C and D: product <6 and sum >5

From list, when sum>5: sums 6,7,8

With product<6: for sum=6: (2,4)p=8>6 not<6, (3,3)p=9>6, (4,2)p=8>6 — none have product<6

Sum=7: (3,4)p=12>6, (4,3)p=12>6
Sum=8: (4,4)p=16>6

So no overlap, P(C and D)=0

Thus P(C or D) = 8/16 + 6/16 - 0 = 14/16 = 7/8

an) Probability that the first spin is greater than 1 and the second spin is less than 3.

First >1: 2,3,4
Second <3: 1,2

So pairs: first in {2,3,4}, second in {1,2}

Number: 3 choices for first, 2 for second, 6 ways: (2,1),(2,2),(3,1),(3,2),(4,1),(4,2)

P=6/16=3/8

ao) Probability that the sum is divisible by 2 or by 3.

Divisible by 2 or 3 means divisible by 2 or 3 or both, i.e., not coprime to 6, but better to use inclusion.

P(div by 2 or 3) = P(div2) + P(div3) - P(div6)

Div by 2: sum even, from ac) 8 ways, P=1/2

Div by 3: sum=3,6,9 but 9 not possible, sum=3 or 6

Sum=3:2 ways, sum=6:3 ways, total 5 ways, P=5/16

Div by 6: sum=6, since 12 too big, sum=6:3 ways, P=3/16

So P = 8/16 + 5/16 - 3/16 = 10/16 = 5/8

List to verify: sums div by 2 or 3: all sums except those not div by 2 or 3, i.e., sum=5,7

Sum=5:4 ways, sum=7:2 ways, total 6 ways not div by 2 or 3? Sum=5 not div by 2 or 3? 5 div by 5, not 2 or 3.

Sum=7 not div by 2 or 3.

Sum=2: div by 2, sum=3: div by 3, sum=4: div by 2, sum=5: not, sum=6: div by 2 and 3, sum=7: not, sum=8: div by 2.

So not div by 2 or 3: sum=5 and sum=7, 4+2=6 ways

So div by 2 or 3: 16-6=10 ways, P=10/16=5/8

ap) Probability that the product is a prime number.

Prime products: 2,3

From grid: (1,2)=2, (2,1)=2, (1,3)=3, (3,1)=3

Also (2,1)=2, etc.

So (1,2),(2,1),(1,3),(3,1) —4 ways

P=4/16=1/4

aq) Probability that the first spin is a square number.

Square numbers: 1,4 (1=1^2, 4=2^2, 2 and 3 not squares)

So first spin 1 or 4, P=2/4=1/2

ar) Probability that the sum is less than 6 and the product is greater than 4.

Sum<6 and product>4.

Sum<6: sum=2,3,4,5

Product>4: 6,8,9,12,16

But sum<6, so possible sums 2,3,4,5 with product>4.

From list:

Sum=2: (1,1)p=1<4 no
Sum=3: (1,2)p=2<4, (2,1)p=2<4 no
Sum=4: (1,3)p=3<4, (2,2)p=4 not >4, (3,1)p=3<4 no
Sum=5: (1,4)p=4 not >4, (2,3)p=6>4, (3,2)p=6>4, (4,1)p=4 not >4

So only (2,3) and (3,2) satisfy: sum=5<6, product=6>4

Two ways.

P=2/16=1/8

as) Probability that the first spin is odd and the second spin is even.

Similar to n), P= (2/4)*(2/4)=1/4

List: (1,2),(1,4),(3,2),(3,4) —4 ways, P=4/16=1/4

at) Probability that the product is less than or equal to 8.

Products ≤8: from grid, exclude >8: 9,12,16

Products >8: (3,3)=9, (3,4)=12, (4,3)=12, (4,4)=16 —4 ways

So P=1 - 4/16 = 12/16 = 3/4

au) Probability that the sum is greater than 3 and less than 7.

Sum>3 and <7, so sum=4,5,6

Sum=4:3 ways, sum=5:4 ways, sum=6:3 ways, total 3+4+3=10 ways

P=10/16=5/8

av) Probability that the first spin is a multiple of the second spin.

Similar to ag), but reverse.

If second spin divides first spin.

List:

Second=1: divides all: (1,1),(2,1),(3,1),(4,1) —4 ways
Second=2: divides 2,4: (2,2),(4,2) —2 ways
Second=3: divides 3: (3,3) —1 way
Second=4: divides 4: (4,4) —1 way

Total 4+2+1+1=8 ways, same as ag), P=8/16=1/2

aw) Probability that the product is even or the sum is even.

P(even product or even sum)

But even sum implies both same parity, even product implies at least one even.

Note that if sum is even, then both even or both odd; if both odd, product odd; if both even, product even.

If sum is odd, then one even one odd, product even.

So the only case where product is odd and sum is odd is when both odd.

Otherwise, either product even or sum even or both.

P(neither) = P(product odd and sum odd) = P(both odd) = (2/4)*(2/4)=1/4

So P(at least one) = 1 - 1/4 = 3/4

P=12/16=3/4

ax) Probability that the first spin is greater than the second spin and the product is even.

First > second and product even.

From earlier, first > second: 6 cases: (2,1),(3,1),(3,2),(4,1),(4,2),(4,3)

Now which have product even.

(2,1):p=2e, (3,1):p=3o, (3,2):p=6e, (4,1):p=4e, (4,2):p=8e, (4,3):p=12e

So only (3,1) has odd product, others even.

So 5 out of 6 have even product.

Cases: (2,1)e, (3,1)o, (3,2)e, (4,1)e, (4,2)e, (4,3)e — so five even product.

P=5/16

ay) Probability that the sum is a multiple of 4.

Sum div by 4: sum=4,8

Sum=4: (1,3),(2,2),(3,1) —3 ways
Sum=8: (4,4) —1 way
Total 4 ways

P=4/16=1/4

az) Probability that the product is greater than 4 and less than 10.

Product >4 and <10: 6,8,9

From grid: (2,3)=6, (2,4)=8, (3,2)=6, (3,3)=9, (4,2)=8

Also (3,2)=6, etc.

List: (2,3):6, (2,4):8, (3,2):6, (3,3):9, (4,2):8

That's five ways.

Is (4,3)=12>10 no, (3,4)=12>10 no.

So five.

P=5/16

ba) Probability that the first spin is even and the sum is odd.

First even and sum odd.

Sum odd implies one even one odd, so if first even, second must be odd.

Which is exactly what we had in n): first even and second odd, P=1/4

And in that case, sum is odd, as established.

So P=1/4

bb) Probability that the product is a multiple of 4.

Multiple of 4: 4,8,12,16

From grid:

(1,4)=4, (2,2)=4, (2,4)=8, (4,1)=4, (4,2)=8, (4,3)=12, (4,4)=16, (3,4)=12

Also (2,2)=4, (1,4)=4, (4,1)=4, (2,4)=8, (4,2)=8, (3,4)=12, (4,3)=12, (4,4)=16

List:
(1,4):4
(2,2):4
(2,4):8
(4,1):4
(4,2):8
(3,4):12
(4,3):12
(4,4):16

That's eight ways.

Is (3,4) included, yes.

P=8/16=1/2

bc) Probability that the sum is less than 5 or greater than 7.

Sum<5 or sum>7.

Sum<5: sum=2,3,4 — from earlier, sum=2:1, sum=3:2, sum=4:3, total 6 ways
Sum>7: sum=8 — (4,4) —1 way
No overlap, so total 6+1=7 ways

P=7/16

bd) Probability that the first spin is a prime number or the second spin is a composite number.

P(prime first or composite second)

Prime first: 2,3 — P=2/4=1/2
Composite second: 4 — P=1/4
P(and) = P(first prime and second composite) = as in ai) 2/16=1/8

So P(or) = 1/2 + 1/4 - 1/8 = 4/8 + 2/8 - 1/8 = 5/8

List: first prime: (2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4) —8 ways
Second composite: second=4: (1,4),(2,4),(3,4),(4,4) —4 ways
Overlap: first prime and second=4: (2,4),(3,4) —2 ways

Union: 8 + 4 - 2 = 10 ways

P=10/16=5/8

be) Probability that the product is less than 6 and the sum is greater than 4.

Product<6 and sum>4.

Product<6: 1,2,3,4
Sum>4: 5,6,7,8

From list, when sum>4 and product<6.

Sum=5: (1,4)p=4<6, (2,3)p=6 not<6, (3,2)p=6 not, (4,1)p=4<6 — so (1,4) and (4,1)

Sum=6: all products >=6, not <6
Sum=7,8: products larger.

So only (1,4) and (4,1)

Two ways.

P=2/16=1/8

bf) Probability that the first spin is odd and the product is even.

First odd and product even.

First odd: 1,3
Product even: requires at least one even, so since first is odd, second must be even.

So second even: 2,4

P= P(first odd) * P(second even) = (2/4)*(2/4) =1/4

List: (1,2),(1,4),(3,2),(3,4) —4 ways, P=4/16=1/4

bg) Probability that the sum is a perfect cube.

Perfect cubes: 1,8

Sum=1 impossible, sum=8: (4,4) —1 way

P=1/16

bh) Probability that the product is greater than the sum or they are equal.

From al), we had p>s:8 ways, p=s:1 way ( (2,2) ), so total 9 ways

P=9/16

bi) Probability that the first spin is greater than 2 or the second spin is less than 2.

First >2: 3,4 — P=2/4=1/2
Second <2: 1 — P=1/4
P(and) = P(first>2 and second<2) = P(first in {3,4}, second=1) = (2/4)*(1/4) =2/16=1/8

P(or) = 1/2 + 1/4 - 1/8 = 4/8 + 2/8 - 1/8 = 5/8

List: first>2: (3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4) —8 ways
Second<2: second=1: (1,1),(2,1),(3,1),(4,1) —4 ways
Overlap: first>2 and second=1: (3,1),(4,1) —2 ways

Union: 8+4-2=10 ways, P=10/16=5/8

bj) Probability that the product is a multiple of 2 and the sum is a multiple of 3.

Product multiple of 2: even product, which is when not both odd, P=3/4, but we need intersection with sum multiple of 3.

Sum multiple of 3: sum=3,6 —5 ways as before.

Among these, which have even product.

Sum=3: (1,2)p=2e, (2,1)p=2e — both even product
Sum=6: (2,4)p=8e, (3,3)p=9o, (4,2)p=8e — so (2,4) and (4,2) have even product, (3,3) has odd.

So for sum=3: both have even product
For sum=6: two out of three have even product

Total: 2 (from sum=3) + 2 (from sum=6) =4 ways

List: (1,2),(2,1),(2,4),(4,2)

P=4/16=1/4

bk) Probability that the first spin is a factor of 4 or the second spin is a multiple of 2.

Factor of 4: 1,2,4 (since 1|4,2|4,4|4, 3 does not)

Multiple of 2: 2,4

P(first factor of 4) = P(1,2,4) =3/4
P(second multiple of 2) = P(2,4) =2/4=1/2
P(and) = P(first in {1,2,4} and second in {2,4}) = (3/4)*(1/2) =3/8? No, independent, but let's calculate number.

First factor of 4: 1,2,4 — three choices
Second multiple of 2: 2,4 — two choices
But for intersection, both conditions.

Number of ways: first in {1,2,4}, second in {2,4}: 3*2=6 ways: (1,2),(1,4),(2,2),(2,4),(4,2),(4,4)

P(and) =6/16=3/8

P(or) = P(A) + P(B) - P(A and B) = (12/16) + (8/16) - (6/16) = 14/16 = 7/8

P(A) = first factor of 4: first=1,2,4 — 3/4 of 16=12 ways? First spin has 4 options, each with 4 second, so for first=1,2,4: 3*4=12 ways, yes P=12/16=3/4

P(B) = second multiple of 2: second=2 or 4, for each first, so 4*2=8 ways, P=8/16=1/2

P(A and B) = as above, 6 ways, P=6/16=3/8

So P(or) = 12/16 + 8/16 - 6/16 = 14/16 = 7/8

bl) Probability that the sum is less than 4 or the product is greater than 8.

Sum<4: sum=2,3 — (1,1),(1,2),(2,1) —3 ways
Product>8: (3,3)=9, (3,4)=12, (4,3)=12, (4,4)=16 —4 ways
No overlap, since sum<4 implies small numbers, product small.

So total 3+4=7 ways

P=7/16

bm) Probability that the first spin is even and the second spin is a prime number.

First even: 2,4
Second prime: 2,3

P= P(first even) * P(second prime) = (2/4)*(2/4) =1/4

List: (2,2),(2,3),(4,2),(4,3) —4 ways, P=4/16=1/4

bn) Probability that the product is less than 5 or the sum is greater than 6.

Product<5: 1,2,3,4 — from c) 8 ways
Sum>6: sum=7,8 — (3,4),(4,3),(4,4) —3 ways
Overlap: product<5 and sum>6 — impossible since sum>6 requires larger numbers, product at least for (3,4)=12>5, so no overlap.

P=8/16 + 3/16 =11/16

bo) Probability that the first spin is greater than the second spin and the sum is even.

First > second and sum even.

Sum even implies both same parity.

First > second.

From earlier first > second: 6 cases: (2,1),(3,1),(3,2),(4,1),(4,2),(4,3)

Sum even: both even or both odd.

(2,1):2e,1o — different parity, sum odd
(3,1):3o,1o — both odd, sum even
(3,2):3o,2e — different, sum odd
(4,1):4e,1o — different, sum odd
(4,2):4e,2e — both even, sum even
(4,3):4e,3o — different, sum odd

So only (3,1) and (4,2) have sum even.

(3,1): sum=4 even, (4,2): sum=6 even

Two ways.

P=2/16=1/8

bp) Probability that the product is a perfect square and the sum is odd.

Perfect square product: from o) 6 ways: (1,1),(1,4),(2,2),(3,3),(4,1),(4,4)

Sum odd: for these, (1,1)s=2e, (1,4)s=5o, (2,2)s=4e, (3,3)s=6e, (4,1)s=5o, (4,4)s=8e

So sum odd for (1,4) and (4,1)

Two ways.

P=2/16=1/8

bq) Probability that the first spin is a multiple of 3 or the second spin is a multiple of 3.

Multiple of 3: 3

P(first=3) =1/4
P(second=3) =1/4
P(and) = P(both=3) =1/16

P(or) = 1/4 + 1/4 - 1/16 = 8/16 + 8/16 - 1/16 = 15/16

List: first=3: (3,1),(3,2),(3,3),(3,4) —4 ways
Second=3: (1,3),(2,3),(3,3),(4,3) —4 ways
Overlap: (3,3) —1 way

Union: 4+4-1=7 ways? 7/16, but 15/16 is wrong.

Mistake.

P(first=3) = number of ways first=3: 4 ways (since second can be anything), so 4/16=1/4

Similarly P(second=3) =4/16=1/4

P(both=3) =1/16

P(or) = 1/4 + 1/4 - 1/16 = 4/16 + 4/16 - 1/16 = 7/16

Yes, 7 ways: all where first=3 or second=3: (3,1),(3,2),(3,3),(3,4),(1,3),(2,3),(4,3) —7 ways, yes.

P=7/16

br) Probability that the sum is less than 6 and the product is less than 6.

Sum<6 and product<6.

Sum<6: sum=2,3,4,5
Product<6: 1,2,3,4

From list, when both.

Sum=2: (1,1)p=1<6
Sum=3: (1,2)p=2<6, (2,1)p=2<6
Sum=4: (1,3)p=3<6, (2,2)p=4<6, (3,1)p=3<6
Sum=5: (1,4)p=4<6, (2,3)p=6 not<6, (3,2)p=6 not, (4,1)p=4<6

So all except (2,3) and (3,2) for sum=5.

So for sum=2:1 way, sum=3:2 ways, sum=4:3 ways, sum=5: (1,4),(4,1) —2 ways (since (2,3),(3,2) excluded)

Total: 1+2+3+2=8 ways

P=8/16=1/2

bs) Probability that the first spin is odd and the sum is greater than 4.

First odd: 1,3
Sum>4: 5,6,7,8

P= P(first odd) * P(sum>4 | first odd) but better list.

First=1: sum>4: second=4 (sum=5), second=3 sum=4 not>4, second=2 sum=3, second=1 sum=2 — so only (1,4) sum=5>4

First=3: sum>4: second=2 sum=5>4, second=3 sum=6>4, second=4 sum=7>4, second=1 sum=4 not>4 — so (3,2),(3,3),(3,4)

So cases: (1,4), (3,2),(3,3),(3,4) —4 ways