Question

Suppose that 10% of all students at Cooley High play a winter sport, that 20% play a spring sport, and that 5% play both a winter and a spring sport. What is the probability that a randomly chosen student plays a winter or a spring sport?
a. 30%
b. 25%
c. 35%
d. 20%
e. None of the above

Answer 1

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:

$A = a + (A \cap B)$

In which a is the probability that a student plays a winter sport both not a spring sport and $A \cap B$ is the probability that a student plays both a winter and a spring sport.

By the same logic, we have that:

$B = b + (A \cap B)$

5% play both a winter and a spring sport.

This means that $A \cap B = 0.05$

20% play a spring sport

This means that $B = 0.2$

10% of all students at Cooley High play a winter sport

This means that $A = 0.1$

What is the probability that a randomly chosen student plays a winter or a spring sport?

$(A \cup B) = A + B - (A \cap B) = 0.2 + 0.1 - 0.05 = 0.25$

So the correct answer is:

b. 25%

Answer 2

Answer:

$P_A = 0.1 , P_B = 0.20 , P(A \cap B) = 0.05$

And for this case we want to calculate this probability:

$P(A \cup B)$

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:

$P(A \cup B) = P(A) +P(B) -P(A \cap B)$

And replacing we got:

$P(A \cup B) = 0.1 +0.2 -0.05 = 0.25$

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

$A \cap B$= A student selected play both a winter and a spring sport

And we have the following probabilities associated to the events:

$P_A = 0.1 , P_B = 0.20 , P(A \cap B) = 0.05$

And for this case we want to calculate this probability:

$P(A \cup B)$

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:

$P(A \cup B) = P(A) +P(B) -P(A \cap B)$

And replacing we got:

$P(A \cup B) = 0.1 +0.2 -0.05 = 0.25$

So then the best option for this case would be:

b. 25%