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Four-teachers-A-B-C-and-D-each-proposed-two-exercises-one-on-algebra-and-another-on-analyses-to-form-an-exam-The-students-have-to-choose-two-exercises-at-random-1-Calculate-the-probability-P-a

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Question Number 110911 by Ar Brandon last updated on 31/Aug/20
Four teachers A, B, C, and D each proposed two exercises, one on algebra and another on analyses, to form an exam. The students have to choose two exercises at random. 1. Calculate the probability P(a) of a student to choose two exercises on algebra. a\ P(a)=(3/(16)) , b\P(a)=(3/(14)) , c\P(a)=(1/4) , d\None 2. Calculate the probability P(b) of choosing two exercises proposed by the same teacher. a\P(b)=(1/(10)) , b\P(b)=(1/(60)) , c\P(b)=(1/7) , d\None 3. Calculate the probability P(c) of choosing two exercises proposed by teacher A. a\P(c)=(1/3) , b\P(c)=(1/4) , c\P(c)=(1/(28)) , d\None
$$\mathrm{Four}\:\mathrm{teachers}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C},\:\mathrm{and}\:\mathrm{D}\:\mathrm{each}\:\mathrm{proposed}\:\mathrm{two}\:\mathrm{exercises}, \\ $$$$\mathrm{one}\:\mathrm{on}\:\mathrm{algebra}\:\mathrm{and}\:\mathrm{another}\:\mathrm{on}\:\mathrm{analyses},\:\mathrm{to}\:\mathrm{form}\:\mathrm{an}\:\mathrm{exam}. \\ $$$$\mathrm{The}\:\mathrm{students}\:\mathrm{have}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{two}\:\mathrm{exercises}\:\mathrm{at}\:\mathrm{random}. \\ $$$$\mathrm{1}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{P}\left(\mathrm{a}\right)\:\mathrm{of}\:\mathrm{a}\:\mathrm{student}\:\mathrm{to}\:\mathrm{choose} \\ $$$$\mathrm{two}\:\mathrm{exercises}\:\mathrm{on}\:\mathrm{algebra}. \\ $$$$\mathrm{a}\backslash\:\mathrm{P}\left(\mathrm{a}\right)=\frac{\mathrm{3}}{\mathrm{16}}\:,\:\mathrm{b}\backslash\mathrm{P}\left(\mathrm{a}\right)=\frac{\mathrm{3}}{\mathrm{14}}\:,\:\mathrm{c}\backslash\mathrm{P}\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:,\:\mathrm{d}\backslash\mathrm{None} \\ $$$$\mathrm{2}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{P}\left(\mathrm{b}\right)\:\mathrm{of}\:\mathrm{choosing}\:\mathrm{two}\:\mathrm{exercises} \\ $$$$\mathrm{proposed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{same}\:\mathrm{teacher}. \\ $$$$\mathrm{a}\backslash\mathrm{P}\left(\mathrm{b}\right)=\frac{\mathrm{1}}{\mathrm{10}}\:,\:\mathrm{b}\backslash\mathrm{P}\left(\mathrm{b}\right)=\frac{\mathrm{1}}{\mathrm{60}}\:,\:\mathrm{c}\backslash\mathrm{P}\left(\mathrm{b}\right)=\frac{\mathrm{1}}{\mathrm{7}}\:,\:\mathrm{d}\backslash\mathrm{None} \\ $$$$\mathrm{3}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{P}\left(\mathrm{c}\right)\:\mathrm{of}\:\mathrm{choosing}\:\mathrm{two}\:\mathrm{exercises} \\ $$$$\mathrm{proposed}\:\mathrm{by}\:\mathrm{teacher}\:\mathrm{A}. \\ $$$$\mathrm{a}\backslash\mathrm{P}\left(\mathrm{c}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:,\:\mathrm{b}\backslash\mathrm{P}\left(\mathrm{c}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:,\:\mathrm{c}\backslash\mathrm{P}\left(\mathrm{c}\right)=\frac{\mathrm{1}}{\mathrm{28}}\:,\:\mathrm{d}\backslash\mathrm{None} \\ $$

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