- 07/04/2019 at 8:19 am #1325461EduGorillaKeymasterSelect Question Language :
Total no. of possible outcomes = 12C3 = 220. Number of ways of selecting three defective bulbs from 6 defective bulbs = 6C3=20. Probability that the room is not lighted = 20/220=1/11. Probability that the room is lighted= 1-1/11=10/11.
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