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There are $40\%$ science students and the remaining are $60\%$ are arts students. It is known that $5\%$ of science students are girls and $10\%$ of the arts students are girls. One student selected at random is a girl. What is probability that she is an arts student?

A girl is randomly selected and there are total $15\%$ girls out of which $10\%$ of girls are from the arts stream, so the probability that she is an arts student$=\frac{10}{15}=\frac{2}{3}$, but it's wrong. Please help.

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  • $\begingroup$ That's a very poor title up there! $\endgroup$ Mar 26, 2015 at 8:11

2 Answers 2

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Let $G$ represent the event "a girl is selected" and let $A$ be the event "an arts student is selected". Then, obviously, $\neg G$ is the event "A boy is selected" and $\neg A$ is the event "a science student is selected".

What you need to calculate is the probability

$$P(A|G)$$

Using Bayes's formula, you know that $$P(G|A) = =\frac{P(G|A)\cdot P(A)}{P(G)}$$

Now, what are the missing numbers? Well, $P(G|A)$ is the probability of selecting a girl if you know you selected an arts student, meaning that's $0.1$ ($10\%$). $P(A)$ is the probability of selecting an arts student, that's $0.6$, i.e. $60\%$.

The probability that a random student is a girl is trickier, but you can use total probability:

$$P(G) = P(G|A)\cdot P(A) + P(G|\neg A)\cdot P(\neg A)$$

There are no unknown numbers in this expression, so $P(G)$ is easy to calculate.

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$2\%$ of students are female science students (because $5\%$ of $40\%$, or $1/20$ of $40\%$, is $2\%$) and $6\%$ of students are female arts students (because $10\%$ of $60\%$, or $1/10$ of $60\%$, is $6\%$). Thus, six of every eight female students are arts students, so $75\%$.

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