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Q.

A lot of 60 bulbs contain 12 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in first attempt is defective and is not replaced. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?


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a

3 5 , 47 59  

b

1 5 , 47 59  

c

1 5 , 48 59  

d

3 5 , 48 59   

answer is C.

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Detailed Solution

As it is given that a lot of 60 bulbs contain 12 defective ones. One bulb is drawn at random from the lot.
Let us find the probability that this bulb is defective.
Total number of outcomes is = 60
Number of defective bulbs is 12 .
So, number of favourable outcomes = 12
The formula of the probability of an event is the number of favourable outcomes divided by the total number of outcomes.
P(E)= no. of favorable outcomes  total  number of outcomes   
Probability that the bulb is defective is
  P = 12 60 P = 1 5  
In the other case, the bulb is drawn in first attempt and not replaced, so the total number of outcomes is 601=59   .
In first attempt, a defective bulb is drawn and not replaced, so the number of favourable outcomes is 5911=48   Probability that the bulb is not defective in second attempt is
P= 48 59  
Thus, Probability that the bulb is defective is 1 5   , the probability that the bulb is not defective in second attempt is 48 59   .
Hence, the correct option is 3.
 
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