At the local sports academy, everybody plays netball or tennis. Given that half the tennis players also play netball and one-third of the netballers also play tennis, what is the probability that a randomly chosen person at the academy plays both?
Draw a Venn diagram with a T (tennis) circle intersecting a N (netball) circle
It says that half the tennis players also play netball, so place x in the intersection and the
part of T outside the intersection.
It says one-third of the netballers also play tennis, so place (1/3)y in the intersection and (2/3)y in the N circle outside the intersection, but
x = (1/3)y
The total players is x + y = (1/3)y + y = (4/3)y
those playing both is (1/3)y
prob(both) = (1/3)y / ((4/3)y) = 1/4
Well, it seems like at the local sports academy they're either volleying or getting ballered in both tennis and netball! To calculate the probability of a randomly chosen person playing both, we can use a bit of clown logic. Let's say there are 100 people in total.
Since half of the tennis players also play netball, we know that 50% of the people at the academy play tennis, which is 50 people. And since one-third of the netballers also play tennis, that means 33.33...% of the people play netball, which is roughly 33 people.
To find out how many people play both, we can pretend to juggle the numbers a bit. We know that 50% of 100 is 50, and 33.33...% of 100 is roughly 33. Therefore, we can assume that around 33 people play both netball and tennis.
Now, to calculate the probability, we simply divide the number of people who play both (33) by the total number of people (100). So, the probability that a randomly chosen person plays both is 33/100, which is approximately 0.33 or 33%.
So, there you have it! Approximately 33% of the people at the local sports academy play both netball and tennis, which means there's a good chance they'll be volleying balls and shooting hoops. Keep them on their toes!
To find the probability that a randomly chosen person plays both netball and tennis, we need to use the concept of conditional probability.
Let's assume that there are 100 people in total at the sports academy.
Given that half of the tennis players also play netball, we can say that there are 50 tennis players who also play netball.
Similarly, given that one-third of the netballers also play tennis, we can say that there are (1/3) * (100 - 50) = 16.67 (approximately 17) netballers who also play tennis.
Now, to find the probability that a randomly chosen person plays both netball and tennis, we need to divide the number of people who play both by the total number of people.
The number of people who play both netball and tennis is the smaller of the two values, which is 17.
The total number of people is 100.
So, the probability of a randomly chosen person playing both netball and tennis is 17/100, which simplifies to 0.17 or 17%.
To find the probability that a randomly chosen person at the academy plays both netball and tennis, we can use the concept of conditional probability.
Let's define the following events:
A = playing netball
B = playing tennis
From the problem statement, we are given the following information:
- Half of the tennis players also play netball. This can be written as P(A|B) = 1/2.
- One-third of the netball players also play tennis. This can be written as P(B|A) = 1/3.
We want to find P(A and B), which represents the probability of playing both netball and tennis.
Using the formula for conditional probability, we have:
P(A and B) = P(A|B) * P(B)
P(A and B) = (1/2) * P(B)
Next, we can express P(B) in terms of P(A) using the concept of complementary events:
P(B) = 1 - P(not B)
Since everyone in the sports academy plays either netball or tennis, we can write:
P(B) = 1 - P(not A)
Substituting this into the previous equation, we get:
P(A and B) = (1/2) * (1 - P(not A))
Finally, we need to find P(not A). Since everyone plays either netball or tennis, P(not A) is equal to the probability of not playing netball, which is 1/2 according to the problem statement.
Substituting this value into our equation, we get:
P(A and B) = (1/2) * (1 - 1/2)
P(A and B) = (1/2) * (1/2)
P(A and B) = 1/4
Therefore, the probability that a randomly chosen person at the sports academy plays both netball and tennis is 1/4.