A group of yr11 students aked to choose subjects for yr12. 135 chose mathematics subject (m) 84 chose language (L), 55 chose both. Find probability Pr(M) Pr(L) Pr(MnL)you can use a venn diagram and...

A group of yr11 students aked to choose subjects for yr12. 135 chose mathematics subject (m) 84 chose language (L), 55 chose both.

Find probability

Pr(M)

Pr(L)

Pr(MnL)

you can use a venn diagram and also find pr(MuL)

 

1 Answer | Add Yours

mlehuzzah's profile pic

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

 

You don't mention how many people chose neither. So I am going to say that there n students who chose neither.

There are 135 students who chose math, and 55 who chose both, so there are 135-55 = 80 students who chose math but didn't choose language

There are 84 students who chose language, and 55 who chose both, so there are 84-55=29 students who chose language but not math.

So we have:

n students who chose neither
80 students who chose math and not language
29 students who chose language and not math
55 students who chose both language and math
----
164 students taking math or language (or both)
164+n total students

`Pr(M)` = math students / total students = `135/(164+n)`
`Pr(L)` = language students / total students = `84/(164+n)`
`Pr(M uu L)` = (students taking math or language or both) / total students = `164/(164+n)`
`Pr(M nn L)` = (students taking neither) / total students = `n/(164+n)`

So for example, if there were 10 students taking neither:

`Pr(M) = 135/174`
`Pr(L) = 84/174`
`Pr(M uu L) = 164/174`
`Pr(M nn L) = 10/174`

 

 

Or, if there were 0 students taking neither (which might be your scenario, but I didn't want to assume)

`Pr(M) = 135/164`
`Pr(L) = 84/164`
`Pr(M uu L) = 164/164=1`
`Pr(M nn L) = 0/164=0`

 

 

 

We’ve answered 318,959 questions. We can answer yours, too.

Ask a question