Question

at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of the students participate in either sports or drama, what is the probability that a student participates in both sports and drama?

Answer 1

P( A Student participates in Both sports and Drama) =0.07

What is the probability?

The probability of an occurrence can be determined using the probability formula by simply dividing the favorable number of possibilities by the entire number of possible outcomes.

To find the probability that a student participates in both sports and drama:

Let

A =Student  participate in Sport

B = Student  participate in Drama

Thus we have:

P(A) = 0.42

P(B) = 0.25

and

P(Ac and B or A and Bc) =0.53

P(Ac and B ) + P( A and Bc) =0.53

P(B) - P(A and B) + P(A) - P(A and B) = 0.53

0.25 - P(A and B )  +  0.42 - P(A and B) = 0.53

0.25+0.42 - 2 * P( A and B) = 0.53

0.67 - 2 * P(A and B) = 0.53

0.67 - 0.53  = 2* P(A and B)

0.14  =  2 * P(A and B)

0.14 / 2  = P( A and B)

0.07 = P( A and B)

that is:

P( A and B) = 0.07

P( A Student  participate in Both sports and Drama) =0.07.

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Answer 2

The probability that a student participates in both sports and drama is $0.14$.

What is the formula for P(AUB), where A and B are any two events?

If $A$ and $B$ are any two events, then the probability of the joint event $(A\cup B)$ is given by the following formula: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose $S$ denotes that "a student participates in sports" and $D$ denotes that "a student participates in drama".
So, we have $P(S)=\frac{42}{100}=0.42$, $P(D)=\frac{25}{100}=0.25$, $P(S\cup D)=\frac{53}{100}=0.53$.

We want to find the probability that a student participates in both sports and drama i.e., we want to find $P(S\cap D)$.

By the above formula, we obtain:

$P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}$

Therefore, the probability that a student participates in both sports and drama is $0.14$.

To learn more about probability, refer: brainly.com/question/24756209

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