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got it
cubic terms are also there to simplify. i dint find any simplification of cubic terms in the text book? and how?
so i shall remind u tht today is the day u have been waiting for to ans my questions..............
ill post the booklet from other threads which has martices transformation notes. i hope u find it helpful
i solved it this way and i got the answer. but my concern is the 25 in the brackets. if we are taking x out as common we know that 25 is not multiplied by x originally so how come ?
i do not understand wht it is tht concerns you.......when we take x common 25x becomes 25 ......
but 25 x is not there in the equation ...its only 25.. so how is x common to all terms of equation
A(a) the probablity that one is faulty is 1/20 given so p is 1/20 and the probability that its not is the 19 parts left of 20 so 19/20 is q
(c) now the third pick should be also one of the faulty after the probability that the two first picked have one faulty too. this is part bii * 1/20 (p. of third faulty) i hope you understood.
OMG i dint notice
silly me . never mind and thanks a lot
this is with replacment.......and in the question it is written EACH time a faulty calculator is chosed the probability is 1/20....there u goView attachment 38825A(a) the probablity that one is faulty is 1/20 given so p is 1/20 and the probability that its not is the 19 parts left of 20 so 19/20 is q
(bi) probability that two chosen are faulty is going to be 1/20 * 1/20 that is 1/400
(bii) exactly one is faulty could be either the first one or the second so = 1/20 * 19/20 + 19/20* 1/20 = 38/400 = 19/200View attachment 38826(c) now the third pick should be also one of the faulty after the probability that the two first picked have one faulty too. this is part bii * 1/20 (p. of third faulty)
(d) the probability that both faulty is 1/20 * 1/20 already calculated =1/400 and the third being faulty is 19/4000 that is part c add both and you'll get the answer.
(e) this is the simplest. you already got part d the probability that the stock gets rejected. multiply it with 1000 to get the numbers of faulty ones!
though i dint understanad why the total amount of calculators remain equal in all cases at the denominator such that these events are mutually exclusive !
walaikum asalamAssalamu Alayikum
ok, wait...
last two parts plss
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