A box contains 15 tube lights, out of which 6 are repair. 3 tube lights are chosen at random from this box. Find the probability that at least one of these is repair
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uhh
15-6=9
Only three chosen
IDK
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Given :
Total Tube lights = 15
No. of repair Tube lights = 6
No. of randomly chosen Tube lights = 3
P(none is repair) = 9C3 / 15C3
= (9*8*7 / 3*2*1) / (15*14*13 / 3*2*1)
= (504/6) / (2730 / 6)
= 504/6 * 6/2730
= 504/2730
= 12/65
P(at least one is repair) = 1 – 12/65
= 53/65
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