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A-number-is-selected-at-random-from-the-integers-10-to-35-exclussive-such-that-the-chances-take-the-subsets-of-P-Q-and-R-respectively-where-P-is-even-Q-is-even-or-prime-number-and-R-is-odd-prime




Question Number 11599 by tawa last updated on 28/Mar/17
A number is selected at random from the integers 10 to 35 exclussive, such  that the chances take the subsets of P, Q and R respectively, where P is  even , Q is even or prime number, and R is odd prime numbers. What is  the probability   (a) That exacly one of the events of the subset is choosen ?  (b) Of choosen at most two of the subset of the event.
$$\mathrm{A}\:\mathrm{number}\:\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random}\:\mathrm{from}\:\mathrm{the}\:\mathrm{integers}\:\mathrm{10}\:\mathrm{to}\:\mathrm{35}\:\mathrm{exclussive},\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{chances}\:\mathrm{take}\:\mathrm{the}\:\mathrm{subsets}\:\mathrm{of}\:\mathrm{P},\:\mathrm{Q}\:\mathrm{and}\:\mathrm{R}\:\mathrm{respectively},\:\mathrm{where}\:\mathrm{P}\:\mathrm{is} \\ $$$$\mathrm{even}\:,\:\mathrm{Q}\:\mathrm{is}\:\mathrm{even}\:\mathrm{or}\:\mathrm{prime}\:\mathrm{number},\:\mathrm{and}\:\mathrm{R}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{prime}\:\mathrm{numbers}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{probability}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{That}\:\mathrm{exacly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{events}\:\mathrm{of}\:\mathrm{the}\:\mathrm{subset}\:\mathrm{is}\:\mathrm{choosen}\:? \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Of}\:\mathrm{choosen}\:\mathrm{at}\:\mathrm{most}\:\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{subset}\:\mathrm{of}\:\mathrm{the}\:\mathrm{event}. \\ $$
Answered by mrW1 last updated on 29/Mar/17
10,11,...,35⇒36 pieces  P=(10,12,...,34)⇒13 pieces  Pr=(11,13,17,19,23,31)⇒6 pieces  Q=P or Pr⇒13+6=19 pieces  R=Pr⇒6 pieces  (a) (6/(36))=16.6%  (b) ((13)/(36))=36.1%
$$\mathrm{10},\mathrm{11},…,\mathrm{35}\Rightarrow\mathrm{36}\:{pieces} \\ $$$${P}=\left(\mathrm{10},\mathrm{12},…,\mathrm{34}\right)\Rightarrow\mathrm{13}\:{pieces} \\ $$$${Pr}=\left(\mathrm{11},\mathrm{13},\mathrm{17},\mathrm{19},\mathrm{23},\mathrm{31}\right)\Rightarrow\mathrm{6}\:{pieces} \\ $$$${Q}={P}\:{or}\:{Pr}\Rightarrow\mathrm{13}+\mathrm{6}=\mathrm{19}\:{pieces} \\ $$$${R}={Pr}\Rightarrow\mathrm{6}\:{pieces} \\ $$$$\left({a}\right)\:\frac{\mathrm{6}}{\mathrm{36}}=\mathrm{16}.\mathrm{6\%} \\ $$$$\left({b}\right)\:\frac{\mathrm{13}}{\mathrm{36}}=\mathrm{36}.\mathrm{1\%} \\ $$
Commented by tawa last updated on 29/Mar/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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