Question

Out of 30 bulbs 6 are defective if 4 bulbs are choosen at random find the probability o f choosing 2 defective bulbls if replacement is done?

Answer 1

@satellite73 @eliassaab

Answer 2

two in a row?

Answer 3

probability first is defective is \(\frac{6}{30}=\frac{1}{5}\) so probability two are defective (with replacement) is \(\frac{1}{5}\times \frac{1}{5}\)

Answer 4

@satellite73 my quewstion was wrong plzz read it now

Answer 5

@helder_edwin

Answer 6

@RadEn

Answer 7

2 defective with probability \(\frac{1}{5}\) each, 2 not defective with probability \(\frac{4}{5}\) each you have
\[P(x=2)=\binom{4}{2}\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^2\]