Respuesta :
Answer:
b. 25%
Step-by-step explanation:
I am going to say that:
A is the probability that a student plays a winter sport.
B is the probability that a student plays a spring sport.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a student plays a winter sport both not a spring sport and [tex]A \cap B[/tex] is the probability that a student plays both a winter and a spring sport.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
5% play both a winter and a spring sport.
This means that [tex]A \cap B = 0.05[/tex]
20% play a spring sport
This means that [tex]B = 0.2[/tex]
10% of all students at Cooley High play a winter sport
This means that [tex]A = 0.1[/tex]
What is the probability that a randomly chosen student plays a winter or a spring sport?
[tex](A \cup B) = A + B - (A \cap B) = 0.2 + 0.1 - 0.05 = 0.25[/tex]
So the correct answer is:
b. 25%
Answer:
[tex] P_A = 0.1 , P_B = 0.20 , P(A \cap B) = 0.05[/tex]
And for this case we want to calculate this probability:
[tex] P(A \cup B) [/tex]
Who represent the probability that a randomly chosen student plays a winter or a spring sport
And for this case we can use the total rule of probability given by:
[tex] P(A \cup B) = P(A) +P(B) -P(A \cap B)[/tex]
And replacing we got:
[tex] P(A \cup B) = 0.1 +0.2 -0.05 = 0.25[/tex]
So then the best option for this case would be:
b. 25%
Step-by-step explanation:
For this case we define the following events:
A = A student selected play a winter sport
B= A student selected play a spring sport
[tex]A \cap B[/tex]= A student selected play both a winter and a spring sport
And we have the following probabilities associated to the events:
[tex] P_A = 0.1 , P_B = 0.20 , P(A \cap B) = 0.05[/tex]
And for this case we want to calculate this probability:
[tex] P(A \cup B) [/tex]
Who represent the probability that a randomly chosen student plays a winter or a spring sport
And for this case we can use the total rule of probability given by:
[tex] P(A \cup B) = P(A) +P(B) -P(A \cap B)[/tex]
And replacing we got:
[tex] P(A \cup B) = 0.1 +0.2 -0.05 = 0.25[/tex]
So then the best option for this case would be:
b. 25%
