[sel1234]

A survey of the reading habits of some students revealed that, on a regular basis, 25% read quality newspapers, 45% read tabloid newspapers and 40% do not read newspapers at all.

a Find the proportion of students who read both quality and tabloid newspapers.

Can someone please explain where they get 0.6 from ?

heres the working;
P(Q∪T)= P(Q)+ P(T)− P(Q∩T)

0.6 =0.25+0.45− P(Q∩T)

How did 'they' work out P(Q∪T)= 0.6 from the information given ?

[Rosh_123]

Ok: Adding up 25, 45 and 40 gives 110. So, 10% must read both, giving 15% reading only quality newspapers, 35% reading only tabloids and 10% reading both and 40% reading none. P(UuT) is therefore 35% + 15% + 10% = 60% = 0.6.

Hope that helps!

[ghostwalker]

If 40% do not read a paper, then 60% (0.6) read one, or other, or both.

[sel1234]

Originally Posted by ghostwalker

If 40% do not read a paper, then 60% (0.6) read one, or other, or both.

so its suppose to be out of 200%?

[ghostwalker]

Originally Posted by sel1234

so its suppose to be out of 200%?

NO.


P(QuT)=probability that they read one (a quality), or the other (a tabloid), or both (a tabloid and a quality) = 0.6

[vc94]

A venn diagram makes the solution easier.

[Prokaryotic_crap]

P(Q u T) + P(no newspapers) = 1

P(Q u T) = 1 - P(no newspapers) = 1 - 0.4 = 0.6

i just sone a similar question lol

[Prokaryotic_crap]

Originally Posted by Rosh_123

Ok: Adding up 25, 45 and 40 gives 110. So, 10% must read both, giving 15% reading only quality newspapers, 35% reading only tabloids and 10% reading both and 40% reading none. P(UuT) is therefore 35% + 15% + 10% = 60% = 0.6.

Hope that helps!

i think thats abit confusing, well for me anyway lol

ghostwalkers way is better