Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items, find (a)P{X = 0} and (b) P {X > 2}.

Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items, find (a)P{X = 0} and (b) P {X > 2}.

Question
Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items, find (a)P{X = 0} and (b) P {X > 2}.

Answers (1)

2021-01-20
The event { X=0 } is the event in which 10 non-defective are drawn one after another.
It would go down like this:
The probability of the1st item being non-defective is 94/100,because there are 94 non defective items and 100 items total.
The probability of picking the 2nd non-defective is 93/99, because there are now only 93 non-defective items left, and only a total of 99 items left (we took one out, remember?).
The probability of picking the 3rd non-defective items is 92/98,and so on until all 10 items have been picked.
Each of these are independent events, so we can just multiply them together to get the final result:
\(\displaystyle{\left({\frac{{{94}}}{{{100}}}}\right)}{\left({\frac{{{93}}}{{{99}}}}{\left({\frac{{{92}}}{{{98}}}}\right)}{\left({\frac{{{91}}}{{{97}}}}\right)}{\left({\frac{{{90}}}{{{96}}}}\right)}{\left({\frac{{{89}}}{{{95}}}}\right)}{\left({\frac{{{88}}}{{{94}}}}\right)}{\left({\frac{{{87}}}{{{93}}}}\right)}{\left({\frac{{{86}}}{{{92}}}}\right)}{\left({\frac{{{85}}}{{{91}}}}\right)}={\left({\frac{{{32808912827897606400}}}{{{62815650955529472000}}}}\right)}\approx{0.522}\right.}\)
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