A box contains 100 bulbs 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that: i. all 10 are defective ii. all 10 are good ii. at least one is defective iv. none3 is defective.

Explanation

total no of bulbs `=100`.

total no of defective bulbs `=20`.
we have take out 10 bulbs, therefore total possible outcomes is `.^100C_10`
1. if all the 10 are defective means taken out from 20 defective bulbs i.e possible outcome is `.^20C_10`
therefore probability is` p= (.^20C_10) /( .^100C_10)`
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`
therefore robability is ` p= (.^80C_10) /( .^100C_10)`
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)`
therefore probability of atleast one defective is ` 1-p` = `1-(.^20C_10) /( .^100C_10)`