Can someone help with these

Mathematics

1 answer:

Dvinal [7]3 years ago

6 0

Answer:

i dont answer your question

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3 years ago

at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t

baherus [9]

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

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

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:

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