A light bulb will be selected at random from the light bulbs at this store. Find the probability that the selected light bulb is fluorescent given that it is labeled as long life. Round to 2 decimal places.

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

x = 25/2304

step-by-step explanation:

6 hours 48 min

step-by-step explanation:

68÷5=13.6

13.6×30=408min=6 hours 48 min.

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