Let's find hte prob that none are bad.
P(0) = 40/50*39/49*38/48 *37/47*36/46
= 0.3106
P(1 bad) = 5 * 40/50*39/49*38/48 *37/47*10/46
= 0.4313
P(0 or 1) = 0.7419
2007-06-12 11:06:21
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answer #1
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answered by Dr D 7
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Because the lot size (50) is finite, the appropriate probability distribution is the hypergeometric distribution. The formula for this is:
p(x) = (DCx)(N-DCn-x/NCn
Where C means Combination
D is the number of defects in the lot (10)
N is the size of the lot (50)
n is the sample size (5)
x is the number of defectives in the sample (either 0 or 1)
Evaluating,
P(x = 0) = (10C0)(40C5)/(50C5) = (1)(658008)/2118760 = 0.31056
P(x = 1) =(10C1) (40C4)/(50C5) = (10)(91390)/2118760 = 0.43134
And the probability (x ⤠1) = P(x = 0) + P(x = 1) = 0.74190
2007-06-12 19:00:33
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answer #2
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answered by cvandy2 6
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The probability of your question can be stated as:
When you take out 1 bulb at random, the prob of gettin a good one is 40/50 = 4/5 and that of getting a bad one is 10/50= 1/5. Hence, the sequence of taking out 5 bulbs out randomly should be :
G,G,G,G,G OR
G,G,G,G,B OR
G,G,G,B,B OR
G,G,B,B,B OR
G,B,B,B,B OR
B,B,B,B,B. Where G = good bulb B = bad bulb
Now acc to ur question,
the situation is :G,G,G,G,B. (for at least one bad bulb)
so, the probability for atleast on bad bulb should be:
5/6
( since there are 6 trials and there is atleast on bad bulb in 5 trials)
2007-06-12 18:41:24
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answer #3
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answered by rickomak f 2
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The sample : all combination C(50,5) = 50!/(5!)(45!)
The event: combination with at least a defective bulb
If you put the 10 defectives aside and evaluate C(40,5) you will find all combinations without defectives
All combinations - combinations without defectives = combinations with at least one defective
Then the number of elements in the event is:
C(50,5) - C(40,5)
And the probability is P(E) = |E| / |S| =
[ C(50,5) - C(40,5)]/ C(50,5)
2007-06-12 18:08:11
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answer #4
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answered by vahucel 6
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