Math, asked by ashmeet21131, 6 months ago

In a group of 20,000 men and 10,000 women, 6 per cent of the men and 3 per cent of the women have
contained affliction. What is the probability that an afflicted member of the group is a man?

Answers

Answered by aadishjain71
0

Step-by-step explanation:

This involves using the Law of Total Probability and Bayes' Theorem. Let + and - represent positive and negative tests; let D and N represent having the disease and not. Here is an outline to get you started. Try to match each part to formulas in your text.

You seek P(D|+)=P(D∩+)/P(+).P(D|+)=P(D∩+)/P(+).

For the numerator, use P(D∩+)=P(D)P(+|D),P(D∩+)=P(D)P(+|D), where numerical values for both factors are given in the statement of the problem.

For the denominator, start with P(+)=P(D∩+)+P(N∩+).P(+)=P(D∩+)+P(N∩+). You already know the first term from the numerator. Find a similar way to evaluate the second term from information given in the problem.

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