Q.It's given that 80% of people attend their family and doctor regularly; 35% of these people have no health problems cropping up during the following year. Out of the 20% of people who don't see their doctor regularly, only 5% have no health issues during the following year. What is the probability a person selected at random will have no health problems in the following year?

Conditional Probability: The "Given That" Idea

Imagine you're rolling a fair six-sided die. The probability of getting a 4 is $\frac{1}{6}$ — straightforward.

Now suppose I roll the die, you don't see it, but I tell you: "The number is even." Given that piece of information, what's the probability that it's a 4? Suddenly, the only possible outcomes are 2, 4, and 6. Among these three equally likely cases, exactly one is a 4. So the probability becomes $\frac{1}{3}$.

That's conditional probability in action: the probability of one event, recalculated under the knowledge that another event has already occurred.

The Intuition: Shrinking the Sample Space

When you condition on an event, you're not changing the die or the randomness. You're simply restricting your attention to a smaller set of possibilities — the part of the original sample space where the condition holds.

In the die example:

• Original sample space: $\{1,2,3,4,5,6\}$ (size 6)
• Condition "even": $\{2,4,6\}$ (size 3)
• Favorable outcome "4": $\{4\}$ (size 1)
• Conditional probability = $\frac{1}{3}$

The key shift: you're no longer asking "out of all possible outcomes, how many are favorable?" You're asking "out of only those outcomes that satisfy the condition, how many are favorable?"

The Precise Definition

Let $A$ and $B$ be two events, with $P(B) > 0$. The conditional probability of $A$ given $B$ is:

Read "$P(A \mid B)$" as "probability of A given B."

Why this formula? Because:

• $P(A \cap B)$ counts outcomes where both A and B happen — these are the only ones that matter once B is known.
• $P(B)$ is the total probability of the condition — we divide by it to rescale the probabilities so that the condition's total probability becomes 1.

In the die example: $A = \{4\}$, $B = \{2,4,6\}$.

• $P(A \cap B) = P(\{4\}) = \frac{1}{6}$
• $P(B) = \frac{3}{6} = \frac{1}{2}$
• $P(A \mid B) = \frac{1/6}{1/2} = \frac{1}{3}$

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