MHT CET · Maths · Probability
A lot of 100 bulbs contains 10 defective bulbs. Five bulbs selected at random from the lot and sent to retail store, then the probability that the store will receive at most one defective bulb is
- A 0.59049
- B 0.91854
- C 0.6561
- D 0.32805
Answer & Solution
Correct Answer
(B) 0.91854
Step-by-step Solution
Detailed explanation
Probability of finding defective bulb \(=\frac{10}{100}=0.1\)
Let \(\mathrm{p}=0.1 \Rightarrow 1=0.9\)
We have \(\mathrm{n}=5\) and to get required probability \(\mathrm{x}=0\) and \(\mathrm{x}=1\)
\(
\begin{aligned}
& \therefore \mathrm{P}={ }^5 \mathrm{C}_0(0.1)^2(0.9)^5+{ }^5 \mathrm{C}_1(0.1)^1(0.9)^4 \\
& (0.9)^5+(5)(0.1)(0.9)^4 \\
& =(0.9)^4[0.9+0.5]=(0.9)_4(1.4) \\
& =(0.81)(0.81)(1.4)=(0.6561)(1.4)=0.91854
\end{aligned}
\)
Let \(\mathrm{p}=0.1 \Rightarrow 1=0.9\)
We have \(\mathrm{n}=5\) and to get required probability \(\mathrm{x}=0\) and \(\mathrm{x}=1\)
\(
\begin{aligned}
& \therefore \mathrm{P}={ }^5 \mathrm{C}_0(0.1)^2(0.9)^5+{ }^5 \mathrm{C}_1(0.1)^1(0.9)^4 \\
& (0.9)^5+(5)(0.1)(0.9)^4 \\
& =(0.9)^4[0.9+0.5]=(0.9)_4(1.4) \\
& =(0.81)(0.81)(1.4)=(0.6561)(1.4)=0.91854
\end{aligned}
\)
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