Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027
<h3>
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➷ I'm assuming you are asking for the lowest common multiple
[if not, let me know]
The answer would be 3000
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➶ Hope This Helps You!
➶ Good Luck (:
➶ Have A Great Day ^-^
↬ ʜᴀɴɴᴀʜ ♡
Answer: 4.8
Step-by-step explanation:
Corresponding sides of similar figures are proportional, so
Answer:
x=9
Step-by-step explanation:
