Q119 / 160 · WBJEE 2015 · Mathematics

Suppose a machine produces metal parts that contain some defective parts witl probability $0.05 .$ How many parts should be produced in order that the probability 0 atleast one part being defective is $1 / 20$ more? (Given that, $\log _{10} 95=1.977$ an $\log _{10} 2=0.3$ )

  • A
    11
  • B
    12
  • Correct answer
    C
    15
  • Correct answer
    D
    14

Solution

Given probability of delective pat $=0.05=\frac{1}{20}$ Probability of non-defective part $=1-0.05=0.95=\frac{19}{20}$ We know that, $P(X=r)={ }^{n} Cr p^{r} q^{n-r}$ where, $\quad p=\frac{1}{20}, q=\frac{19}{20}$ $r \geq 1$ and $n=?$ Also, $P(X \geq 1) \geq \frac{1}{2}$ $\Rightarrow 1-P(X=0) \geq \frac{1}{2}$ $\Rightarrow 1-{ }^{n} C_{0}\left(\frac{1}{20}\right)^{0}\left(\frac{19}{20}\right)^{n-0} \geq \frac{1}{2}$ $\Rightarrow \quad 1-\frac{1}{2} \geq\left(\frac{19}{20}\right)^{n}$ $\Rightarrow$ $\frac{1}{2} \geq\left(\frac{19}{20}\right)^{n}$ $\Rightarrow$ $\frac{1}{2} \geq\left(\frac{95}{100}\right)^{n}$ $\Rightarrow \quad \log 2^{-1} \geq n[\log 95-\log 100 \mid$ $\Rightarrow \quad-\log 2 \geq n[\log 95-2]$ $\Rightarrow \quad-(0.3) \geq n[1977-2]$ $\Rightarrow \quad n \geq \frac{03}{0.023}$ $\Rightarrow \quad n \geq \frac{300}{23}$ $\Rightarrow \quad n \geq 1304$ $\therefore \quad n=14,15$

Hence, the correct answer is options: 'C', 'D'.

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