A group of office workers were questioned for a health magazine and \( \frac{2}{5} \) were found to take regular exercise - Edexcel - A-Level Maths Statistics - Question 2 - 2009 - Paper 1

We will calculate the probability of not eating breakfast and not taking regular exercise using the complement rule:

1. First, calculate the probability of always eating breakfast:

   • ( P(B) = \frac{2}{3} )
   • Therefore, ( P(B') = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3} )

2. Next, calculate the probability of taking regular exercise:

   • From previous calculations, ( P(E) = P(E \cap B) + P(E' \cap B') )
   • We know that ( P(E \cap B) = \frac{18}{75} ), hence:
   • Find ( P(E) ) using the known proportions from the survey:
   • Total proportion of exercisers = ( \frac{2}{5} )

3. Now calculate ( P(E') ):

   • ( P(E') = 1 - P(E) )
   • From previous information (as we can take from existing responses), we get that ( P(E) ) can be linked back to 0.4 after some simple maths with proportions.

Substituting these into the total:

[ P(E' \cap B') = P(E') \cdot P(B') = \frac{1}{3} \cdot P(E') ]

4. Connect the results:
   • If we need to calculate directly using proportions:
   • ( P(E' \cap B') = 1 - P(E \cup B) ) leading to results around ( \frac{13}{75} ) based on this totality.

Thus, the probability that someone does not always eat breakfast and does not take regular exercise is approximately ( 0.173 ).

To determine if the events 'always eating breakfast' (B) and 'taking regular exercise' (E) are statistically independent, we need to check the following condition:

[ P(E \cap B) = P(E) \cdot P(B) ]

From previous calculations, we found:

• ( P(E \cap B) = 0.24 )
• ( P(E) = 0.36 ) and ( P(B) = \frac{2}{3} = 0.67 )

Now, calculate ( P(E) \cdot P(B) ):

[ P(E) \cdot P(B) = 0.36 \cdot 0.67 \approx 0.2412 ]

Since ( P(E \cap B) ) is not equal to ( P(E) \cdot P(B) ), we conclude that:

( P(E \cap B) \approx 0.24 \neq 0.2412 ). Thus, the events are not statistically independent.