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Question Number 193226 by byaw last updated on 07/Jun/23
  (a) 5 out of 12 articles are known to be defective. If three articles are picked, one after the other, without replacement, find the probability that all the three articles are non-defective.     (b) Two coins are tossed and a dice is thrown. What is the probability of obtaining a head, a tail and a 4?
(a) 5 out of 12 articles are known to be defective. If three articles are picked, one after the other, without replacement, find the probability that all the three articles are non-defective.

(b) Two coins are tossed and a dice is thrown. What is the probability of obtaining a head, a tail and a 4?

Answered by MM42 last updated on 10/Jun/23
a) (7/(12))×(6/(11))×(5/(10))=(7/(44))  b)(1/2)×(1/2)×(1/6)=(1/(24))
a)712×611×510=744b)12×12×16=124
Commented by byaw last updated on 09/Jun/23
my dear I do not understand
mydearIdonotunderstand
Commented by MM42 last updated on 10/Jun/23
a)total number of articles = 12=n(s)  number of defective articles=5  ⇒number of non−defective articles=7=n(a)  because every time we select without   inserting one is reduced fromr the   previous number .  first time : n(s)=12  ;  n(a)=7⇒p=(7/(12))  second time : n(s)=11 ; n(a)=6⇒p=(6/(11))  third time ; n(s)=10  ;  n(a)=5⇒p=(6/(10))  ⇒answer=multiplication   b)  each coin has 2 states amd  each  dice has  6 states .  first coin : p_1 =(1/2)  & second  coin : p_2 =(1/2)   &  coin : p_3 =(1/6)  ⇒answer=multiplication
a)totalnumberofarticles=12=n(s)numberofdefectivearticles=5numberofnondefectivearticles=7=n(a)becauseeverytimeweselectwithoutinsertingoneisreducedfromrthepreviousnumber.firsttime:n(s)=12;n(a)=7p=712secondtime:n(s)=11;n(a)=6p=611thirdtime;n(s)=10;n(a)=5p=610answer=multiplicationb)eachcoinhas2statesamdeachdicehas6states.firstcoin:p1=12&secondcoin:p2=12&coin:p3=16answer=multiplication

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