Question

A 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

.78 rounded to the nearest hudredth as asked in the orignal question. You are welcome! :)

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:

Given : A 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.

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

Solution :  

Let B be event to choose boys and G be event to choose girls.

Let E be event to scored grade A.

We have given,

P(B)=55%=0.55

P(G)=45%=0.45

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

$P(E/B)=\dfrac{P(E\cap B)}{P(B)}=\frac{0.25}{0.55}=0.45$

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

$P(E/G)=\dfrac{P(E\cap G)}{P(B)}=\frac{0.35}{0.45}=0.78$

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.

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

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

$P(G/E)=0.5865$

$P(G/E)=58.65\%$

Therefore, 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%