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(C) Fill in the Blanks in the Table: P (A) P (B) P (A ∩ B) P(A∪ B) 0.35 .... 0.25 0.6

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Question

Fill in the blank in the table:

P (A) P (B) P (A ∩ B) P(A∪ B)
0.35 .... 0.25 0.6
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Solution

Given: \[P\left( A \right) = \frac{1}{3}, P\left( B \right) = \frac{1}{5} \text{ and }  P\left( A \cap B \right) = \frac{1}{15}\]

Given:
P (A) = 0.35, P (A ∪ B) = 0.6 and P (A ∩ B) = 0.25
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) - P (A ∩ B)
          0.6  = 0.35 + P (B) - 0.25
       P (B) = 0.6 - 0.35 + 0.25
         = 0.6 - 0.1 = 0.5

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Concept of Probability - Probability of 'Not', 'And' and 'Or' Events
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Chapter 33: Probability - Exercise 33.4 [Page 67]

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R.D. Sharma Mathematics [English] Class 11
Chapter 33 Probability
Exercise 33.4 | Q 1.3 | Page 67

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