Question

b) what if you returned to check on the bin and found that there were 4 teddy bears left and 12 other animals? what could you add to or remove from the bin to return the probability of selecting a teddy bear to ½?

Explanation:

Step1: Define current counts

Let \( t = 4 \) (teddy bears), \( o = 12 \) (other animals). Total \( n = t + o = 4 + 12 = 16 \).

Step2: Let \( x \) be change in teddy bears, \( y \) in others.

Desired probability: \( \frac{t + x}{(t + x) + (o + y)} = \frac{1}{2} \).

Step3: Simplify the equation

Cross - multiply: \( 2(t + x)=(t + x)+(o + y) \) → \( t + x=o + y \).

Step4: Substitute \( t = 4 \), \( o = 12 \)

\( 4 + x=12 + y \) → \( y=x - 8 \).

Step5: Analyze possible integer solutions

• Case 1: Add teddy bears (x > 0, y can be adjusted)

If we set \( y = 0 \) (no change in other animals), then \( x=8 \). So add 8 teddy bears: new \( t = 4+8 = 12 \), \( o = 12 \), total \( 24 \), probability \( \frac{12}{24}=\frac{1}{2} \).

• Case 2: Remove other animals (y < 0, x can be 0)

If \( x = 0 \), then \( y=- 8 \). So remove 8 other animals: new \( o = 12 - 8 = 4 \), \( t = 4 \), total \( 8 \), probability \( \frac{4}{8}=\frac{1}{2} \).

• Case 3: Mixed (add teddy, remove other)

E.g., add 4 teddy (\( x = 4 \)), then \( y=4 - 8=-4 \) (remove 4 other). New \( t = 8 \), \( o = 8 \), total \( 16 \), probability \( \frac{8}{16}=\frac{1}{2} \).

Answer:

One way is to add 8 teddy bears (so teddy bears = 12, other animals = 12, total = 24, probability \( \frac{12}{24}=\frac{1}{2} \)) or remove 8 other animals (teddy bears = 4, other animals = 4, total = 8, probability \( \frac{4}{8}=\frac{1}{2} \)) or other combinations following \( y=x - 8 \). (A common simple answer: Add 8 teddy bears or remove 8 other animals)