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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Probability

NDA 2024 · Mathematics · Probability

The occurrence of a disease in an industry is such that the workers...

The occurrence of a disease in an industry is such that the workers have 20%20 \% chance of suffering from it. What is the probability that out of 6 workers chosen at random, 4 or more will suffer from the disease?

Options

  1. A533125\frac{53}{3125}
  2. B633125\frac{63}{3125}
  3. C733125\frac{73}{3125}
  4. D833125\frac{83}{3125}

Correct answer

A. 533125\frac{53}{3125}

Step-by-step solution

 Here, n=6p=P( Suffering from  disease )=20100=19q=P( Not suffering from disease )=1−15=45P(x≥4)=P(x=4)+P(x=5)+P(x=6)=6c4(15)4(45)2+6c5(15)5(45)1+6c6(15)6=6!4!2!×1656+6!5!×456+156=1(5)6[15×16+6×4+1]=240+24+115625=26.515625=533125\begin{aligned} & \text{ Here, } \mathrm{n}=6 \\ & \mathrm{p}=\mathrm{P}(\text{ Suffering\text{ from } disease })=\frac{20}{100}=\frac{1}{9} \\ & \mathrm{q}=\mathrm{P}(\text{ Not suffering from disease })=1-\frac{1}{5}=\frac{4}{5} \\ & \mathrm{P}(\mathrm{x} \geq 4)=\mathrm{P}(\mathrm{x}=4)+\mathrm{P}(\mathrm{x}=5)+\mathrm{P}(\mathrm{x}=6) \\ & ={ }^6 \mathrm{c}_4\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^2+{ }^6 \mathrm{c}_5\left(\frac{1}{5}\right)^5\left(\frac{4}{5}\right)^1+{ }^6 \mathrm{c}_6\left(\frac{1}{5}\right)^6 \\ & =\frac{6!}{4!2!} \times \frac{16}{5^6}+\frac{6!}{5!} \times \frac{4}{5^6}+\frac{1}{5^6} \\ & =\frac{1}{(5)^6}[15 \times 16+6 \times 4+1] \\ & =\frac{240+24+1}{15625}=\frac{26.5}{15625}=\frac{53}{3125}\end{aligned}
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