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NCERT Class XI Mathematics - Probability - Solutions

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Question : 38 of 54
Marks: +1, -0
If E and F are events such that P(E) = 14\frac{1}{4} , P (F) = 12\frac{1}{2} and P(E and F) = 18\frac{1}{8} , find (i) P(E or F), (ii) P(not E and not F).
Solution:  
(i) P(E or F) = P(E ∪ F)
= P(E) + P(F) – P(E ∩ F) = 14+12−18\frac{1}{4}+\frac{1}{2}-\frac{1}{8} = 2+4−18\frac{2+4-1}{8} = 58\frac{5}{8}
(ii) not E and not F = E′ ∩ F′ = (E ∪ F)′ (De Morgan’s Law)
\ P(not E and not F) = P(E ∪ F)′ = 1 - P (E ∪ F) = 1 - 58\frac{5}{8} = 38\frac{3}{8}
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