S1 probability

Hey, not too sure about this exam question:
2. A group of office workers were questioned for a health magazine and 2/5
were found to take regular exercise. When questioned about their eating habits 2/3said they always eat breakfast and, of those who always eat breakfast 9/25 also took regular exercise.
Find the probability that a randomly selected member of the group
(a) always eats breakfast and takes regular exercise,
I understand that this is 2/3 x 9/25 = 6/25
(b) does not always eat breakfast and does not take regular exercise.
The answer is 13/75, don't really understand how they got that. I think I might be missing the obvious :s-smilie:

If you could help that would be great :smile:

Reply 1

-Matt-

Hello

Basically you need to find out how the proportion that have breakfast and/or take exercise: the result will then be the remainder when this value is subtracted from 1 (as by definition if the proportion remaining neither have breakfast or take exercise). To find the proportion that do either you need to remember that there are some that do both (if thinking of it visually the intersection of two sets on a Venn diagram) - if you add the individual proportions that eat breakfast and that take breakfast, those that do both will be counted twice. Therefore you need to account for this by subtracting this proportion (which you found in part a) from the sum.

If it helps more to have it symbolically:

Let B be the event eats breakfast, E takes exercise

[br]P(\bar{E} \cap \bar{B}) = 1 - P(E \cup B) = 1 - [P(E) + P(B) -P(E \cap B)][br]

Hope this helps,
Matt

Reply 2

Sonia1

OP

-Matt-

Hello

Basically you need to find out how the proportion that have breakfast and/or take exercise: the result will then be the remainder when this value is subtracted from 1 (as by definition if the proportion remaining neither have breakfast or take exercise). To find the proportion that do either you need to remember that there are some that do both (if thinking of it visually the intersection of two sets on a Venn diagram) - if you add the individual proportions that eat breakfast and that take breakfast, those that do both will be counted twice. Therefore you need to account for this by subtracting this proportion (which you found in part a) from the sum.

If it helps more to have it symbolically:

Let B be the event eats breakfast, E takes exercise

[br]P(\bar{E} \cap \bar{B}) = 1 - P(E \cup B) = 1 - [P(E) + P(B) -P(E \cap B)][br]

Hope this helps,
Matt

That did help, thank you! I was pretty confused.