Classroom has 100 students -- probability of speaking a language

[DotKite]

Homework Statement

A large classroom has 100 students. 30 speak spanish, 25 speak italian, and 55 speak neither.

a). how many speak both?

b).A student who speak italian is chosen at random. What is the probability that he/she speaks spanish?

Homework Equations

P(A) = 1 - P(A$^{c}$)
P(U) = 1, where U with the universal set
(A$\cap$B)$^{c}$ = (A$^{c}$ $\cup$ B$^{c}$)

The Attempt at a Solution

so I assume that when they say 30 speak spanish they mean only speak spanish, and the same for italian.

let S be the set of those who speak spanish and let I be the set of those who speak italian.


P(S-I) = 30/100, P(I-S) = 25/100

P(who speak neither) = 55/100

1-P(who speak neither) = P(S $\cup$ I) = 45/100

THen I thought that you would want to do,

P(S $\cup$ I) - [P(S-I) + P(I-S)] in order to get P(S $\cap$ I) but that comes out to be negative so I am clearly not correct. I do not know what else to do. Been on this problem for a ever now.

 

[haruspex]

so I assume that when they say 30 speak spanish they mean only speak spanish, and the same for italian.

Wrong assumption.

 

[DotKite]

ok then i guess you can figure this out by adding all the givens to get 110/100 and deduce that there must be 10 people who speak both spanish and italian. However I was wondering if anyone could show me how to do this with sets? or just a more elegant way?

 

[haruspex]

ok then i guess you can figure this out by adding all the givens to get 110/100 and deduce that there must be 10 people who speak both spanish and italian. However I was wondering if anyone could show me how to do this with sets? or just a more elegant way?

You have |S|=30, |U|=100, etc.
You want |S∩U|.
The key equation to relate cardinality to unions and intersections hinges on the notion of disjoint union: |A| = |A∩B| + |A∩B^c|
Try it from there.

 

[Curious3141]

ok then i guess you can figure this out by adding all the givens to get 110/100 and deduce that there must be 10 people who speak both spanish and italian. However I was wondering if anyone could show me how to do this with sets? or just a more elegant way?

Sketch a Venn diagram.

 

[Ray Vickson]

ok then i guess you can figure this out by adding all the givens to get 110/100 and deduce that there must be 10 people who speak both spanish and italian. However I was wondering if anyone could show me how to do this with sets? or just a more elegant way?

How on Earth can you have 110 people in a group of 100? The fact that you have numbers adding to more than 100 should be enough to tell you that you are on the wrong track. The fact that the question asks you to find the number of people speaking both Spanish and Italian should tell you that some speakers of Spanish also speak Italian, unlike what you assumed.

As others have suggested: draw a Venn diagram.

 

[haruspex]

How on Earth can you have 110 people in a group of 100? The fact that you have numbers adding to more than 100 should be enough to tell you

It may not have been most clearly expressed, but I believe that is what DotKite was saying. By adding up the three numbers and subtracting 100 the number double counted (the bilinguals) is found. What DotKite was looking for next was a more algebraic approach. But that was nearly three years ago.

 

[Ray Vickson]

It may not have been most clearly expressed, but I believe that is what DotKite was saying. By adding up the three numbers and subtracting 100 the number double counted (the bilinguals) is found. What DotKite was looking for next was a more algebraic approach. But that was nearly three years ago.

OK, maybe. I did not notice the "3 years" bit.