Question Detail
A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is
- \begin{aligned} \frac{7}{19} \end{aligned}
- \begin{aligned} \frac{6}{19} \end{aligned}
- \begin{aligned} \frac{5}{19} \end{aligned}
- \begin{aligned} \frac{4}{19} \end{aligned}
Answer: Option A
Explanation:
Please remember that Maximum portability is 1.
So we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs.
So here we go,
Total cases of non defective bulbs
\begin{aligned}
^{16}C_2 = \frac{16*15}{2*1} = 120 \\
\text{total cases = } \\
^{20}C_2 = \frac{20*19}{2*1} = 190 \\
\text{probability = } \frac{120}{190} = \frac{12}{19} \\
\text{P of at least one defective = } 1- \frac{12}{19} \\
=\frac{7}{19}
\end{aligned}
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anu 11 years ago
why we hve taken combination ist ie 16c2
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