Question

if 2% or electric bulbs manufactured by a certain company are deffective . Find the probability that in a sample of 200 bulbs. a) less than 2 bulbs, b) more than 3 bulbs are defective .

Answer 1

In this case the probability of the bulb being defective is very low.

P(X<2)=P(0)+P(1)=

.98^200+200×.98^199×.02=.98^199(.98+4)=4.98×.98^199=.089375+

P(X<4)=P(X<2)+P(2)+P(3)=4.98×.98^199+200C2×.98^198×.02^2+200C3×.98^197×.02^3=4.98×.98^199+(7.8008+10.5072)×.98^197=.089375+.34212—=.43149+. So P(X>3)=1-P(X<4)=.5685+.

Answer 2

Answer:

Probability of manufacturing a defective bulb = 2% = 0.02

Probability of manufacturing a good bulb = 1 - 0.02 = 0.98

Using the formula for binomial distribution :

$\sum_{n=0}^{r}_{r}^{n}\textrm{C}\cdot p^r\cdot q^{n-r}$

Let X be the random variable for manufacturing the defective bulbs

a). n = 200 bulbs and r = 2

$P(X < r = 2) = \sum_{n=0}^{2}_{r}^{200}\textrm{C}\cdot 0.02^r\cdot 0.98^{200-r}\\\\P(X < r = 2) = _{0}^{200}\textrm{C}\cdot 0.02^0\cdot 0.98^{200-0}+_{1}^{200}\textrm{C}\cdot 0.02^1\cdot 0.98^{199}\\\\\implies P(X < r = 2) =0.0894$

b). n = 200 and r = 3

P(X <  4 ) = P(X < 2) + P(2) + P(3)

               = 0.0894 + 0.34212

               = 0.43

⇒ P(X > 3) = 1 - 0.43

                 = 0.57