Saved Bookmarks
| 1. |
Assuming that on an average 1% of the output in a factory making certain part of an article are defective and that 200 units are in a package, find the probabilities that: (i) at most 2 defectives, and (ii) at least two defectives, may be found in the package. |
|
Answer» p = 1% = \(\frac1{100}\) q = 1 - p = 1 - \(\frac1{100}\) = \(\frac{99}{100}\), n = 200 (i) Probability that almost two defective part in the package = \(P(x\le 2)\) \(P(0) + P(1) + P(2)\) \(= \,^{200}C_0 \,p^0\, q^{200} + \,^{200}C_1 \,p\,q^{199}+ \,^{200}C_2 p^2 q^{198}\) \(= \left(\frac1{100}\right)^0 \left(\frac{99}{100}\right)^{200}+ \,^{200}\left(\frac1{100}\right) \left(\frac{99}{100}\right)^{199} + \frac{200.199}{2}\left(\frac1{100}\right)^2\left(\frac{99}{100}\right)^{198}\) \(= \left(\frac{99}{100}\right)^{198} \left(\left(\frac{99}{100}\right)^2 + 2\left(\frac{99}{100}\right)+ \frac{199}{100}\right)\) \(= \left(\frac{99}{100}\right)^{198} \left(\frac{9801 + 19800 + 19900}{10000}\right)\) \(= \left(\frac{99}{100}\right)^{198} \times \left(\frac1{100}\right)^2 \times 49501\) (ii) Probability that at least two defective part in the package = \(P(x\ge 2)\) \(= 1- P(0) - P(1)\) \(= 1- \,^{200}C_0 \,p^0\,q^{200} -\, ^{200}C_1\,p\,q^{199}\) \(= 1 - \left(\frac1{100}\right)^0 \left(\frac{99}{100}\right)-\,^{200}\frac1{100}\left(\frac{99}{100}\right)^{99}\) \(= 1 - \left(\frac{95}{100}\right)^{199} \left(\frac{99}{100}+2\right)\) \(= 1 - \left(\frac{99}{100}\right)^{199} \left(\frac{299}{100}\right)\) |
|