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For the events A and B, P(A ∩ B') = 0.32, P(A' ∩ B) = 0.11 and P(A ∪ B) = 0.65 - Edexcel - A-Level Maths Statistics - Question 6 - 2006 - Paper 1

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For-the-events-A-and-B,--P(A-∩-B')-=-0.32,-P(A'-∩-B)-=-0.11-and-P(A-∪-B)-=-0.65-Edexcel-A-Level Maths Statistics-Question 6-2006-Paper 1.png

For the events A and B, P(A ∩ B') = 0.32, P(A' ∩ B) = 0.11 and P(A ∪ B) = 0.65. (a) Draw a Venn diagram to illustrate the complete sample space for the events A an... show full transcript

Worked Solution & Example Answer:For the events A and B, P(A ∩ B') = 0.32, P(A' ∩ B) = 0.11 and P(A ∪ B) = 0.65 - Edexcel - A-Level Maths Statistics - Question 6 - 2006 - Paper 1

Step 1

Draw a Venn diagram to illustrate the complete sample space for the events A and B.

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Answer

To draw the Venn diagram, we represent events A and B as two overlapping circles within a rectangle which represents the sample space E.

  1. The area for A only is represented as P(A) - P(A ∩ B).
  2. The area for B only is represented as P(B) - P(A ∩ B).
  3. The area where A and B overlap is P(A ∩ B).
  4. The area outside both A and B is represented by the remaining portion of E.

Using the provided probabilities:

  • P(A ∩ B) = 0.32 + 0.11 + 0.35 = 0.54
  • P(A) + P(B) + P(A ∩ B) = 1
  • P(A ∪ B) = 0.65

Calculating further, we find:

  • P(A) = 0.32 + 0.22 = 0.54
  • P(B) = 0.11 + 0.22 = 0.33

Thus the complete diagram will show:

  • A: 0.32
  • B: 0.11
  • A ∩ B: 0.22
  • Outside (E): 0.35.

Step 2

Write down the value of P(A) and the value of P(B).

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Answer

From the calculations in part (a), we find:

  • P(A) = 0.54
  • P(B) = 0.33.

Step 3

Find P(A | B').

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Answer

To find P(A | B'), use the conditional probability formula:

P(AB)=P(AB)P(B)P(A | B') = \frac{P(A \cap B')}{P(B')}

We know:

  • P(A ∩ B') = 0.32
  • P(B') = 1 - P(B) = 1 - 0.33 = 0.67

So,

P(AB)=0.320.670.478P(A | B') = \frac{0.32}{0.67} \approx 0.478

Step 4

Determine whether or not A and B are independent.

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Answer

To determine independence, we check if:

P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

Using our values:

  • P(A ∩ B) = 0.22
  • P(A) = 0.54
  • P(B) = 0.33

Calculating:

P(A)×P(B)=0.54×0.33=0.1782P(A) \times P(B) = 0.54 \times 0.33 = 0.1782

Since P(AB)P(A)×P(B)P(A \cap B) \neq P(A) \times P(B), we conclude that A and B are NOT independent.

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