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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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| \[\frac{1}{3}\] | \[\frac{1}{5}\] | \[\frac{1}{15}\] | ...... |
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| 0.5 | 0.35 | ..... | 0.7 |
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