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Solution :

total no of bulbs=100.<br>
total no of defective bulbs =20.<br>we have take out 10 bulbs, therefore total possible outcomes is `.^100C_10` <br>1. if all the 10 are defective means taken out from 20 defective bulbs i.e possible outcome is `.^20C_10`<br> therefore probability is` p= (.^20C_10) /( .^100C_10)` <br>2.if all the 10 are good that is taken out from rest of the 80 which are good bulbs i.e possible outcome is `.^80C_10`<br> therefore robability is ` p= (.^80C_10) /( .^100C_10)`<br>4.atleast one is defective means we have to find all the probabilities when one is defective ,2 is defective and 3 is defective so on .so find probility when none is defective i.e all are good which is ` p= (.^20C_10) /( .^100C_10)`<br> therefore probability of atleast one defective is ` 1-p` = `1-(.^20C_10) /( .^100C_10)`**What is probability experiments :**

**Deterministic experiments**

**Random or probabilistic experiments**

**Elementary event sample space and examples**

**What is event occurrence of event**

**Algebra of events**

**negation of an event verbal description and equivalent set theoretic notation**

**Certain event and impossible event**

**Compound events**

**Mutually exclusive events**