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Given that P(A) = 0.35 , P(B) = 0.45 and P(A ∩ B) = 0.13 find a) P(A ∪ B) b) P(A' | B') The event C has P(C) = 0.20 - Edexcel - A-Level Maths Statistics - Question 7 - 2013 - Paper 1

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Given-that--P(A)-=-0.35-,-P(B)-=-0.45-and-P(A-∩-B)-=-0.13--find--a)-P(A-∪-B)--b)-P(A'-|-B')--The-event-C-has-P(C)-=-0.20-Edexcel-A-Level Maths Statistics-Question 7-2013-Paper 1.png

Given that P(A) = 0.35 , P(B) = 0.45 and P(A ∩ B) = 0.13 find a) P(A ∪ B) b) P(A' | B') The event C has P(C) = 0.20. The events A and C are mutually exclusive ... show full transcript

Worked Solution & Example Answer:Given that P(A) = 0.35 , P(B) = 0.45 and P(A ∩ B) = 0.13 find a) P(A ∪ B) b) P(A' | B') The event C has P(C) = 0.20 - Edexcel - A-Level Maths Statistics - Question 7 - 2013 - Paper 1

Step 1

a) P(A ∪ B)

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Answer

To find the probability of the union of two events, we can use the formula:

P(AB)=P(A)+P(B)P(AB)P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Substituting the given values:

P(AB)=0.35+0.450.13=0.67P(A ∪ B) = 0.35 + 0.45 - 0.13 = 0.67

Thus, the answer is:

P(A ∪ B) = 0.67

Step 2

b) P(A' | B')

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Answer

To find the conditional probability, we use:

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

First, we need to find P(AB)P(A' ∩ B'):

Using the complement rule:

P(AB)=1P(AB)P(A' ∩ B') = 1 - P(A ∪ B)

Substituting the known value:

P(AB)=10.67=0.33P(A' ∩ B') = 1 - 0.67 = 0.33

Now substituting into the conditional probability formula, where P(B)=1P(B)=0.55P(B') = 1 - P(B) = 0.55:

P(AB)=0.330.550.6P(A' | B') = \frac{0.33}{0.55} \approx 0.6

So, the answer is:

P(A' | B') ≈ 0.6

Step 3

c) Find P(B ∩ C)

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Answer

Since B and C are independent events, the formula for the intersection is:

P(BC)=P(B)imesP(C)P(B ∩ C) = P(B) imes P(C)

Substituting the given values:

P(BC)=0.45imes0.20=0.09P(B ∩ C) = 0.45 imes 0.20 = 0.09

Thus, the answer is:

P(B ∩ C) = 0.09

Step 4

d) Draw a Venn diagram to illustrate the events A, B and C and the probabilities for each region.

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Answer

To draw the Venn diagram, we represent the three events A, B, and C as overlapping circles. The probabilities for each region can be computed as follows:

  1. P(A) = 0.35
  2. P(B) = 0.45
  3. P(C) = 0.20
  4. P(A ∩ B) = 0.13, hence, the intersection of A and B should show 0.13.
  5. Mutually exclusive region for C should be illustrated, which means it does not overlap with A. The remaining probabilities can be calculated based on the total area being 1 with overlaps accounted for. It's essential to also denote P(BC)=0.09P(B ∩ C) = 0.09 from part (c).

The diagram should clearly label each region with the corresponding probabilities.

Step 5

e) Find P([B ∪ C')

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Answer

To find the probability of the union of B and the complement of C, we use:

P(BC)=P(B)+P(C)P(BC)P(B ∪ C') = P(B) + P(C') - P(B ∩ C')

First, calculate P(C)P(C'):

P(C)=1P(C)=10.20=0.80P(C') = 1 - P(C) = 1 - 0.20 = 0.80

Next, we calculate P(BC)P(B ∩ C'). Knowing B and C are independent:

P(BC)=P(B)imesP(C)=0.45imes0.80=0.36P(B ∩ C') = P(B) imes P(C') = 0.45 imes 0.80 = 0.36

Now substitute these into the union formula:

P(BC)=0.45+0.800.36=0.89P(B ∪ C') = 0.45 + 0.80 - 0.36 = 0.89

Thus, the answer is:

P(B ∪ C') = 0.89

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