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C) Fill in the Blanks in the Table: P (A) P (B) P (A ∩ B) P(A∪ B) 1 3 1 5 1 15 ......

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Question

Fill in the blank in the table:

P (A) P (B) P (A ∩ B) P(A∪ B)
\[\frac{1}{3}\] \[\frac{1}{5}\] \[\frac{1}{15}\] ......
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Solution

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

By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) -  P (A ∩ B)

\[= \frac{1}{3} + \frac{1}{5} - \frac{1}{15}\]

\[= \frac{5 + 3 - 1}{15} = \frac{7}{15}\]

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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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