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Question 15 If the probability of the letter chosen at random from the letters of the word "Mathematics" to be a vowel is 2/(2π‘₯+1), then π‘₯ is equal to (A) 4/11 (B) 9/4 (C) 11/4 (D) 4/9Word β€˜MATHEMATICS has total 11 letters M, M, A, A, T, T, H, E, I, C, S Number of letters in word β€˜MATHEMATICS’ = 11 Vowels are a, e, i, o, u Vowels in in word β€˜MATHEMATICS’ is A, A, E, I Number of vowels in word β€˜MATHEMATICS’ = 4 Now, Required Probability = (π‘΅π’–π’Žπ’ƒπ’†π’“ 𝒐𝒇 π’—π’π’˜π’†π’π’” 𝑖𝑛 π‘€π‘œπ‘Ÿπ‘‘ 𝑀𝐴𝑇𝐻𝐸𝑀𝐴𝑇𝐼𝐢𝑆)/(π‘΅π’–π’Žπ’ƒπ’†π’“ 𝒐𝒇 𝒍𝒆𝒕𝒕𝒆𝒓𝒔 𝑖𝑛 π‘€π‘œπ‘Ÿπ‘‘ 𝑀𝐴𝑇𝐻𝐸𝑀𝐴𝑇𝐼𝐢𝑆) = πŸ’/𝟏𝟏 Now, given that Required Probability = 2/(2π‘₯+1) Putting values πŸ’/𝟏𝟏 = 𝟐/(πŸπ’™+𝟏) 4(2x + 1) = 2 Γ— 11 8x + 4 = 22 8x = 22 – 4 8x = 18 x = 18/8 x = πŸ—/πŸ’ So, the correct answer is (b)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo