Conditional Probability Question

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SUMMARY

The probability of selecting two defective televisions from a shipment of 100, where 6 are defective, is calculated as 1/330. The first television has a probability of being defective at 6/100, and the second, given the first is defective, has a probability of 5/99. The correct calculation involves multiplying these probabilities: (6/100) * (5/99) = 1/330. This demonstrates the application of conditional probability in determining the likelihood of multiple dependent events.

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Homework Statement



In a shipment of 100 televisions, 6 are defective. If a person buys two televisions from that shipment, what is the probability that both are defective?

Homework Equations



the Answer is somewhat weird! it says it is 1/330 ! which is really beyond by recognition

The Attempt at a Solution



What I think is that it is 5/99!

The first defective TV would have probability of 6/100 ... but the second defective TV would have probability as 5/99 BECAUSE one is reduced from the defective TV Set and one is also reduced from the sample size.

What do you think?
 
Last edited:
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You are answering the wrong question! Yes, the probability that the second set is also defective is 5/99. But the question asked is "what is the probability that both are defective?"

In order that they both be defective, the first has to already be defective and the probability of that is, as you say, 6/100. The probability that the first is defective and the second is defective is (6/100)(5/99)= (3/50)(5/99)= (1/10)(1/33)= 1/330.

The probability of A and B is (Probability of A) times (Probability of B given A).
Your answer, 5/99, is "Probability of B given A" but you still need to multiply by "Probability of A".
 
wow !

thanks ... you are the best !


one last thing ... does it mean that I am not a complete retard ? :redface:

:)
 

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