Question

A box contains $20$ bulbs of which $3$ are defective. If $3$ bulbs are chosen at ransom then, the probability of having at least one good bulb is:

• A. $\frac{169}{190}$
• B. $\frac{189}{190}$
• C. $\frac{75}{95}$
• D. $\frac{1}{190}$

Answer 1

The correct option is B $\frac{189}{190}$

There are $17$ good bulbs and $3$ defective bulbs.

Probability of having atleast one good bulb in selected $3$ bulbs is equal to one minus probability of having $0$ good bulbs.

Probability of selecting $0$ good bulbs is $\frac{3!}{{{}^{20}C}_3}=\frac{1}{190}$.

Therefore probability of selecting atleast one good bulb is $1-\frac{1}{190}=\frac{189}{190}$.

Therefore the correct option is $B$.