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An experiment consisting of testing calculators one after the other with no replacement until either 2 defective calculators are found or 4 are tested. Find the probability of:

a)the Event E consisting of all outcomes where 1 defective calculator is tested.

I tried analyzing every possible outcome, by looking at every possible combination of defective calculators and successful ones, but I couldn't come up with the right solution. How would I solve the problem?

  • 0
    You have to know the probability of a calculator being defective.2012-12-09

2 Answers 2

1

Let 0 means perfect, 1 means faulty.

The possible outcomes are

$0000,0001$;$0010,0011$ ; $011$; $0100,0101$; $11$;$101$;$1001,1000$;

Among them $0001,0010,0100,1000$ have exactly 1 defective calculator.

So, the required probability will be $\frac4{11}$


If we continue up to $4$ tests the number possible cases are $2^4=16$

Exactly 3 $1$s can occur in $\binom 43=4$ ways.

Exactly 4 $1$s can occur in $\binom 44=1$ way.

So, the number of available cases are $=16-(4+1)=11$

Among $4$ tests, exactly 1 defective calculator can occur in $\binom 41=4$ ways.

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    It's perfect, thank you again!2012-12-09
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Write D for defective, O for OK. The possible outcomes are DD, DOD, DOOD, ODD, ODOD, OODD, OODO, OOOD, OOOO. Presumably what you want to do is figure out the probability of each of these 9 possible outcomes, and add up the ones that belong to your event E.

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    Ahh I see. The book says there are 11 total outcomes, not 9. Is it wrong? My teacher confirmed the book is correct.2012-12-09