Answer:
1) Probability that all five are good = 0.46
2) P(at most 2 defective) = 0.99
3) Pr(at least 1 defective) = 0.54
Step-by-step explanation:
The total number of bulbs = 30
Number of defective bulbs = 4
Number of good bulbs = 30 - 4 = 26
Number of ways of selecting 5 bulbs from 30 bulbs = [tex]30C5 = \frac{30 !}{(30-5)!5!} \\[/tex]
[tex]30C5 = 142506[/tex] ways
Number of ways of selecting 5 good bulbs from 26 bulbs = [tex]26C5 = \frac{26 !}{(26-5)!5!} \\[/tex]
[tex]26C5 = 65780[/tex] ways
Probability that all five are good = 65780/142506
Probability that all five are good = 0.46
2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)
Pr(no defective) has been calculated above = 0.46
Pr(1 defective) = [tex]\frac{26C4 * 4C1}{30C5}[/tex]
Pr(1 defective) = (14950*4)/142506
Pr(1 defective) =0.42
Pr(2 defective) = [tex]\frac{26C3 * 4C2}{30C5}[/tex]
Pr(2 defective) = (2600 *6)/142506
Pr(2 defective) = 0.11
P(at most 2 defective) = 0.46 + 0.42 + 0.11
P(at most 2 defective) = 0.99
3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)
Pr(1 defective) =0.42
Pr(2 defective) = 0.11
Pr(3 defective) = [tex]\frac{26C2 * 4C3}{30C5}[/tex]
Pr(3 defective) = 0.009
Pr(4 defective) = [tex]\frac{26C1 * 4C4}{30C5}[/tex]
Pr(4 defective) = 0.00018
Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018
Pr(at least 1 defective) = 0.54
