Q.Suppose that $6\%$ of the people with blood group O are left handed and $10\%$ of those with other blood groups are left handed. $30\%$ of the people have blood group O. If a left handed person is selected at random, what is the probability that he/she will have blood group O?

Concept: Conditional Probability (Bayes' Theorem)

Let $O$ = event that a person has blood group O, and $L$ = event that a person is left-handed.

We are given:

• $P(O) = 0.30$, so $P(O') = 0.70$
• $P(L \mid O) = 0.06$
• $P(L \mid O') = 0.10$

We need $P(O \mid L)$.

Step 1: Find $P(L)$ using the law of total probability:

Step 2: Apply Bayes' Theorem: …

We start with the idea of conditional probability: we want $P(\text{O} \mid \text{left-handed})$, the probability that a person has blood group O given that they are left-handed. The direct data we have is the other way around — left-handedness given blood group. Bayes’ theorem is the natural tool to reverse the conditioning.

Let’s define events clearly:

• Let $O$ = event that a person has blood group O.
• Let $L$ = event that a person is left-handed.

We are told:

• $P(O) = 0.30$ (30% of people have blood group O).
• $P(L \mid O) = 0.06$ (6% of people with blood group O are left-handed).
• $P(L \mid O^c) = 0.10$ (10% of people with other blood groups are left-handed).

We want $P(O \mid L)$.

1. Find the probability of being left-handed overall, $P(L)$. This is the total probability, which we get by considering both ways a person can be left-handed: either they have blood group O and are left-handed, or they don’t have O and are left-handed.

Using the multiplication rule:

So:

That is, 8.8% of the population is left-handed.

2. Apply Bayes’ theorem. Bayes’ theorem states:

We already have all three pieces:

3. Simplify the fraction. …

Method: Bayes' Theorem with a Two-Group Partition

Use this when a population splits into a group and its complement, each with a known rate of some trait, and you want to reverse the conditioning.

Steps

Step 1: Name the partition and the observed trait.

Let $O$ and $O^{c}$ be the two groups with priors $P(O)$, $P(O^c)=1-P(O)$; let $L$ be the observed trait, with rates $P(L\mid O)$ and $P(L\mid O^c)$.

Step 2: Get the total probability of the trait. …

Common Mistakes

Mistake 1: Putting only one group in the denominator.

Why it's wrong: $P(L)$ must count left-handers from both blood-group O and non-O; using only $P(L\mid O)P(O)$ understates the denominator. Correct approach: use total probability, $P(L)=P(L\mid O)P(O)+P(L\mid O^c)P(O^c)=0.088$.

Mistake 2: Reporting $P(L\mid O)$ when $P(O\mid L)$ is asked. …