| A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that atleast one of these is defective, is |
A) \[\frac{4}{19}\]
B) \[\frac{7}{19}\]
C) \[\frac{12}{19}\]
D) \[\frac{21}{95}\]
Correct Answer: B
Solution :
| P (None is defective) \[=\frac{{}^{16}{{C}_{2}}}{{}^{20}{{C}_{2}}}=\frac{12}{19}\] |
| P (atleast one is defective) \[=\left( 1-\frac{12}{19} \right)=\frac{7}{19}\] |
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