Answer:
0.52
Step-by-step explanation:
The complete question is given as follows:
" Of the light bulbs available at a store, 42% are fluorescent, 23% are labeled as long life, and 12% are fluorescent and long life. "
Solution:-
- We will define two events ( A , B ) as follows:
Event ( A ) : - We randomly select a fluorescent light bulb
Event ( B ) :- We randomly select a light bulb labelled " long life "
- The respective probabilities of occurance of each event are given as:
p ( A ) = 0.42 ( 42 % )
p ( B ) = 0.23 ( 23 % )
- We are asked to compute the conditional probability of picking a fluorescent bulb given that it was labelled " long life ". We can mathematically express the conditional probabilities in terms of events ( A ) and ( B ).
p ( A / B ) = p ( A & B ) / p ( B )
- The numerator of the conditonal probability: p ( A & B ) is the occurence of both event simultaneously that the chosen/picked light bulb is labelled as " long life " and fluroscent. The probability of simultaneous occurance is given as p ( A & B ) = 0.12 ( 12 % ).
- Therefore, the required conditional probability is:
p ( A / B ) = 0.12 / 0.23
p ( A / B ) = 0.52174
- the required answer is to be rounded off to 2 dp. The third dp = 1 < 5. Hence, truncate the 3rd decimal place and on-wards.
- The probability that the selected light bulb is fluorescent given that it is labeled as long life is 0.52
