Question

At a local elementary school, 35 percent of all students have brown eyes, 45 percent have brown hair, and 60 percent have brown hair or brown eyes. A student will be selected at random from the school. Let E represent the event that the selected person has brown eyes, and let H represent the event that the selected person has brown hair. Are E and H mutually exclusive events?

Answer 1

Answer:

No they are not mutually exclusive

Explanation:

We have given that 35 % of the students have brown eyes

So $n(A)=35$ %

And 45 % of the student has brown hair

So $n(B)=45$ %

It is also given that 60 % of the student have both brown eyes and brown hair

So $n(A\cup B)=60$ %

From set rule we know that

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

So $60=45+35-n(A\cap B)$

$n(A\cap B)=20$

So A and B will not be mutually exclusive because for A and B to be mutually exclusive $n(A\cap B)$ must be zero

Answer 2

Answer:

A and B is not mutually exclusive because

$\rm n(A\cap B)\neq 0$

Explanation:

Given :

35% of all students have brown eyes.

45% have brown hair.

60% have brown hair or brown eyes.

Calculation :

Let A be the event where 35% of all students have brown eyes.

So, n(A) = 35%

Let B be the event where 45% have brown hair.

So, n(B) = 45%

The event where 60% have brown hair or brown eyes,

$\rm n(A\cup B)$ = 60%

Through set rule

$\rm n(A\cup B) = n(A)+n(B)-n(A\cap B)$

$\rm n(A\cap B)= 80-60$

$\rm n(A\cap B)$ = 20%

Therefore A and B is not mutually exclusive because

$\rm n(A\cap B)\neq 0$

For more information, refer the link given below

https://brainly.com/question/4546043?referrer=searchResults