In a certain region, 46% of the population is female.It is known that 5% of males and 2% of females are left-handed.A person is chosen at random and found to be left-handed.What is the probability that this person is a male?

from the region is a Female, Male, and, Left handed, resp.

From what is given, we have,

By Definition,

To find the probability that a left-handed person chosen at random is a male, we can use Bayes' theorem.

Let $M$ be the event that the person is male, and $L$ be the event that the person is left-handed.

Given:

• Probability of being male ($P(M)$) = 54% = 0.54
• Probability of being female ($P(F)$) = 46% = 0.46
• Probability of being left-handed given male ($P(L|M)$) = 5% = 0.05
• Probability of being left-handed given female ($P(L|F)$) = 2% = 0.02

We want to find $P(M|L)$, the probability that the person is male given that they are left-handed.

Using Bayes' theorem:

$P(M|L) = \frac{P(L|M) \times P(M)}{P(L)}$

We can find $P(L)$ using the law of total probability:

$P(L) = P(L|M) \times P(M) + P(L|F) \times P(F)$

Substitute the given values:

$P(L) = 0.05 \times 0.54 + 0.02 \times 0.46$

$P(L) = 0.027 + 0.0092$

$P(L) = 0.0362$

Now, substitute the values into Bayes' theorem:

$P(M|L) = \frac{0.05 \times 0.54}{0.0362}$

$P(M|L) = \frac{0.027}{0.0362}$

$P(M|L) \approx 0.7459$

Therefore, the probability that a left-handed person chosen at random is male is approximately $0.7459$, or $74.59\%$.