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February 1, 2015

February 1, 2015

Posted by **Maureen** on Monday, May 6, 2013 at 5:38am.

(i) All 3 are defective

(ii) All 3 are good

(iii) The first selected is defective

(iv) 1 defective and 2 good

(v) 2 defective and 1 good

(vi) At most 2 are good

- Math -
**PsyDAG**, Monday, May 6, 2013 at 1:38pmIf the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

1. (1/5)^3 = ?

2. (4/5)^3 = ?

3. 1/5

4. 1/5 * 4/5 * 4/5 = ?

5. You do it, using similar process.

6. This is none good, one good*or*2 good.

Either-or probabilities are found by adding the individual probabilities.

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