Conditional Probability: Examples and formulas for calculating probabilities of events given certain conditions.

Let’s solve this step by step.

We are given two examples involving probability using tree diagrams and the “Blue Beanie Rule” (which seems to be a made-up name for multiplying probabilities along branches of a tree diagram).

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Example 1:
- Two coins: one is fair (P(heads) = 0.5), the other is biased with P(heads) = 0.7.
- We pick one coin at random → so P(pick fair) = 0.5, P(pick biased) = 0.5.
- Then we flip it once.
- We want P(getting heads).

Tree diagram has:
- First branch: choose coin → Fair (0.5) or Biased (0.5)
- Second branch from each: Heads or Tails

So:
- Path 1: Pick Fair → Get Heads: 0.5 × 0.5 = 0.25
- Path 2: Pick Biased → Get Heads: 0.5 × 0.7 = 0.35

Total P(heads) = 0.25 + 0.35 = 0.6

The text says: “Using the Blue Beanie Rule... gives us p = ½·½ + ½·⅗ = ¼ + ⁷₂₀ = ¹²⁄₂₀ = ³⁄₅ = 0.6” — wait, that doesn’t match our calculation? Let me check:

Wait — in the text it says:
> p = ½·½ + ½·⅗ = ¼ + ⁷⁄₂₀ = ¹²⁄₂₀ = ³⁄₅ = 0.6

But ½·⅗ = 3/10 = 6/20, not 7/20. That must be a typo in the original text.

Actually, if the biased coin has P(heads) = 0.7 = 7/10, then:

½ · 7/10 = 7/20 → yes, that’s correct.

And ½ · ½ = 1/4 = 5/20

So total = 5/20 + 7/20 = 12/20 = 3/5 = 0.6 → Correct.

So Example 1 checks out.

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Example 2:
This is about gender of children.

Assume:
- Each child independently has P(boy) = 0.5, P(girl) = 0.5? But wait — the problem says:

> “Each child born into a family is equally likely to be a boy or a girl.”

BUT then it says:

> “Suppose that in a certain population, the probability that a newborn is a boy is 0.51.”

Ah! So actually, P(boy) = 0.51, P(girl) = 0.49.

Then it asks: What is the probability that a family with three children has exactly two boys?

It shows a tree diagram with 8 paths (since 2^3 = 8 possible outcomes for 3 children).

They list the paths that give exactly two boys:

- BBG: 0.51 × 0.51 × 0.49
- BGB: 0.51 × 0.49 × 0.51
- GBB: 0.49 × 0.51 × 0.51

Note: All three have the same product: (0.51)^2 × 0.49

So total probability = 3 × (0.51)^2 × 0.49

Let’s compute that:

First, 0.51² = 0.2601

Then, 0.2601 × 0.49 = ?

Compute:

0.2601 × 0.49
= 0.2601 × (0.5 - 0.01)
= 0.2601×0.5 - 0.2601×0.01
= 0.13005 - 0.002601
= 0.127449

Then multiply by 3:

3 × 0.127449 = 0.382347

Which rounds to approximately 0.382

The text says:
> “the probability of getting two B’s is 3·(0.51)²·0.49 ≈ 0.382”

Perfect — matches.

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Now, what is the actual question? The user didn’t ask a specific new question — they just uploaded an image showing these two examples.

Looking back at the instructions:
> “Solve the problem accurately.”

But there’s no explicit problem stated beyond the examples. However, since both examples are fully solved in the text, perhaps the intent is to verify or understand them.

Wait — maybe the student is supposed to apply the method to a similar problem? But none is given.

Alternatively, perhaps the task is implicit: to confirm the calculations in Example 2?

Since the only numerical answer requested in the text is for Example 2 (“What is the probability...?”), and it’s already computed as ≈0.382, and we verified it, then that’s likely the final answer expected.

But let’s double-check the arithmetic again carefully:

P(exactly 2 boys in 3 children) = C(3,2) × (0.51)^2 × (0.49)^1
= 3 × 0.2601 × 0.49

Compute 0.2601 × 0.49:

Breakdown:

0.2601 × 0.49
= 2601/10000 × 49/100
= (2601 × 49) / 1,000,000

Compute 2601 × 49:

2601 × 50 = 130,050
Minus 2601 = 130,050 - 2,601 = 127,449

So 127,449 / 1,000,000 = 0.127449

Times 3 = 0.382347 → which is approximately 0.382

Yes.

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Final Answer:
0.382