A sample of 200 households was obtained from a small town. Each household was asked to complete a questionnaire about their purchases of takeaway food. A is the event that a household regularly purchases Indian takeaway food. B is the event that a household regularly purchases Chinese takeaway food. It was observed that $P(B | A) = 0.25$ and $P(A | B) = 0.1$. Of these households, 122 indicated that they did not regularly purchase Indian or Chinese takeaway food. A household is selected at random from those in the sample. Find the probability that the household regularly purchases both Indian and Chinese takeaway food.

A-Level AQA Maths Pure Sequences & Series: A sample of 200 households was o

A sample of 200 households was obtained from a small town - AQA - A-Level Maths Pure - Question 15 - 2019 - Paper 3

Question 15

A sample of 200 households was obtained from a small town. Each household was asked to complete a questionnaire about their purchases of takeaway food. A is the ev... show full transcript

Worked Solution & Example Answer:A sample of 200 households was obtained from a small town - AQA - A-Level Maths Pure - Question 15 - 2019 - Paper 3

Step 1

Find P(A), P(B), and P(A ∩ B)

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Answer

From the problem, we know:

Total households = 200
Households not purchasing Indian or Chinese takeaway = 122 Thus, households purchasing either Indian or Chinese takeaway = 200 - 122 = 78.

Using conditional probabilities:

We have the formula: $P(A | B) = \frac{P(A \cap B)}{P(B)}$ => rearranging gives:

$P(A \cap B) = P(A | B) \times P(B)$

Since $P(B | A) = 0.25$ , we can also derive $P(B)$ as follows: Let $P(A) = x$ and using the fact that: $P(B) = 0.25 * P(A)$ , substituting in the equation, we get:

$x + (0.25 * x) = 78$

Therefore, solving for $x$ gives $P(A) = \frac{78}{1.25} = 62.4$ .

Step 2

Find P(A ∩ B)

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Answer

Now, substituting back the values:

From previous findings, we can calculate $P(B) = 0.25 * 62.4$ = 15.6.
Then use it in the context for $P(A | B)$ :

Therefore, $P(A \cap B) = P(A | B) * P(B) = 0.1 * 15.6 = 1.56.$

Thus, the probability that a randomly selected household regularly purchases both Indian and Chinese takeaway food is:

$P(A \cap B) = \frac{1.56}{200} = 0.0078 \text{ or } 0.78\%.$

A sample of 200 households was obtained from a small town - AQA - A-Level Maths Pure - Question 15 - 2019 - Paper 3

Worked Solution & Example Answer:A sample of 200 households was obtained from a small town - AQA - A-Level Maths Pure - Question 15 - 2019 - Paper 3

Find P(A), P(B), and P(A ∩ B)

Find P(A ∩ B)

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