Question #351021

A lot consists of 10 good articles, 4 with minor defects and 2 with major defects. i. One article is chosen at random. Find the probability that (a) It has no defects, (b) It has no major defects, (c) It is either good or has major defects. ii. Two articles are chosen (without replacement), Find the probability that (a) Both are good (b) Both have major defects (c) At least one is good (d) at most one is good (e) Exactly one is good (f) Neither has major defects  


Expert's answer

1.

(a)


P(0 defects)=1010+4+2=58P(0\ defects)=\dfrac{10}{10+4+2}=\dfrac{5}{8}

(b)


P(0 major defects)=10+410+4+2=78P(0\ major\ defects)=\dfrac{10+4}{10+4+2}=\dfrac{7}{8}

(c)


P(0 defects or major)=10+210+4+2=68P(0\ defects\ or\ major)=\dfrac{10+2}{10+4+2}=\dfrac{6}{8}

2.

(a)


P(2 good)=1016(101161)=38P(2\ good)=\dfrac{10}{16}(\dfrac{10-1}{16-1})=\dfrac{3}{8}

(b)


P(2 major)=216(21161)=1120P(2\ major)=\dfrac{2}{16}(\dfrac{2-1}{16-1})=\dfrac{1}{120}

(c)


P(good1)=1P(0 good)P(good\ge1)=1-P(0\ good)

=1616(61161)=78=1-\dfrac{6}{16}(\dfrac{6-1}{16-1})=\dfrac{7}{8}

(d)


P(good1)=1P(2 good)P(good\le1)=1-P(2\ good)

=138=58=1-\dfrac{3}{8}=\dfrac{5}{8}

(e)


P(exactly 1 good)=1016(6161)+616(10161)P(exactly\ 1\ good)=\dfrac{10}{16}(\dfrac{6}{16-1})+\dfrac{6}{16}(\dfrac{10}{16-1})

=12=\dfrac{1}{2}

(f)


P(0 major)=1416(141161)=91120P(0\ major)=\dfrac{14}{16}(\dfrac{14-1}{16-1})=\dfrac{91}{120}
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