MHT CET · Maths · Probability
Ten bulbs are drawn successively, with replacement, from a lot containing \(10 \%\) defective bulbs, then the probability that there is at least one defective bulb, is
- A \(1-\left(\frac{1}{10}\right)^{10}\)
- B \(1-\left(\frac{3}{10}\right)^{10}\)
- C \(1-\left(\frac{9}{10}\right)^{10}\)
- D \(1-\left(\frac{7}{10}\right)^{10}\)
Answer & Solution
Correct Answer
(C) \(1-\left(\frac{9}{10}\right)^{10}\)
Step-by-step Solution
Detailed explanation
Let X : be the number of defective bulbs
\(\therefore \quad\) Possible values of X is 1 .
Here,
\(\mathrm{n}=\) number of bulbs picked \(=10\).
Let P (probability of getting defective bulb)
\(\begin{aligned}
& =10 \%=\frac{1}{10} \\
\therefore \quad & q=1-p=1-\frac{1}{10}=\frac{9}{10}
\end{aligned}\)
\(\therefore \quad\) Probability that at least one bulb is defective \(=1-\mathrm{P}\) (getting 0 defective bulb)
\(=1-\mathrm{P}(\mathrm{X}=0)\)
\(=1-{ }^{10} \mathrm{C}_0(\mathrm{p})^0(\mathrm{q})^{10-0}\)
\(=1-1 \times 1 \times\left(\frac{9}{10}\right)^{10}\)
\(=1-\left(\frac{9}{10}\right)^{10}\)
\(\therefore \quad\) Possible values of X is 1 .
Here,
\(\mathrm{n}=\) number of bulbs picked \(=10\).
Let P (probability of getting defective bulb)
\(\begin{aligned}
& =10 \%=\frac{1}{10} \\
\therefore \quad & q=1-p=1-\frac{1}{10}=\frac{9}{10}
\end{aligned}\)
\(\therefore \quad\) Probability that at least one bulb is defective \(=1-\mathrm{P}\) (getting 0 defective bulb)
\(=1-\mathrm{P}(\mathrm{X}=0)\)
\(=1-{ }^{10} \mathrm{C}_0(\mathrm{p})^0(\mathrm{q})^{10-0}\)
\(=1-1 \times 1 \times\left(\frac{9}{10}\right)^{10}\)
\(=1-\left(\frac{9}{10}\right)^{10}\)
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