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NCERT Class XI Mathematics - Probability - Solutions

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Question : 36 of 54
Marks: +1, -0
Fill in the blanks in following table :
S.No P (A) P (B) P (A ∩ B) P (A ∪ B)
(i) 13\frac{1}{3}15\frac{1}{5}115\frac{1}{15}...
(ii) 0.35 ... 0.25 0.6
(iii) 0.5 0.35 ...0.7
Solution:  
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 13+15−115\frac{1}{3}+\frac{1}{5}-\frac{1}{15} = 5+3−115\frac{5+3-1}{15} = 715\frac{7}{15}
(ii) 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 = 0.5
(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ 0.7 = 0.5 + 0.35 – P(A ∩ B)
∴ P(A ∩ B) = 0.5 + 0.35 – 0.7 = 0.15
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