Respuesta :
.78 rounded to the nearest hudredth as asked in the orignal question. You are welcome! :)
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
[tex]P(E/B)=\dfrac{P(E\cap B)}{P(B)}=\frac{0.25}{0.55}=0.45[/tex]
35% of the class are girls and scored an A on the test
[tex]P(E/G)=\dfrac{P(E\cap G)}{P(B)}=\frac{0.35}{0.45}=0.78[/tex]
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.
[tex]P(G/E)=\dfrac{P(E/G)\times P(G)}{P(E/G)\times P(G)+P(E/B)\times P(B)}[/tex]
[tex]P(G/E)=\dfrac{0.78\times 0.45}{0.78\times 0.45+0.45\times 0.55}[/tex]
[tex]P(G/E)=0.5865[/tex]
[tex]P(G/E)=58.65\%[/tex]
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%
