Question

class consists of 55% boys and 45% girls. It is observed that 25% of the class are boys and scored an A on the test, and 35% of the class are girls and scored an A on the test. If a student is chosen at random and is found to be a girl, the probability that the student scored an A is

Answer 1

If you add 55% and 45% the total is 100%   so 45/100 are girls.  35% of the girls scored and A  so 35/45  is the probability that it is a girl chosen or 7/9. the probability is 7 out of 9

Answer 2

Answer:

A student is chosen at random and is found to be a girl, the probability that the student scored an A is 0.5865 or 58.65%

Step-by-step explanation:

Let B be event to choose boys.

Let G be event to choose girls.

Let E be event to scored grade A.

In a class consists of 55% boys

Therefore, P(B)=0.55

In a class consists of 45% girls

Therefore, P(G)=0.45

25% of the class are boys and scored an A on the test

Therefore, $P(E/B)=\dfrac{P(E\cap B)}{P(B)}=\frac{0.25}{.55}=.45$

35% of the class are girls and scored an A on the test

Therefore, $P(E/G)=\dfrac{P(E\cap G)}{P(B)}=\frac{0.35}{.45}=.78$

Using Baye's Theorem of probability

If a student is chosen at random and is found to be a girl, the probability that the student scored an A

$P(G/E)=\dfrac{P(E/G)\cdot P(G)}{P(E/G)\cdot P(G)+P(E/B)\cdot P(B)}$

$P(G/E)=\dfrac{0.78\times 0.45}{0.78\times 0.45+0.45\times 0.55}$

$P(G/E)=0.5865$

Thus,  a student is chosen at random and is found to be a girl, the probability that the student scored an A is 0.5865 or 58.65%