Search any question & find its solution
Question:
Answered & Verified by Expert
In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb after selection being put back in the box before making the second selection. The probability that both the bulbs are without defect is
Options:
Solution:
2433 Upvotes
Verified Answer
The correct answer is:
$\frac{16}{25}$
Here $P$ (without defected) $=\frac{8}{10}=\frac{4}{5}=p$
$P_{\text {(defected) }}=\frac{2}{10}=\frac{1}{5}=q$ and $n=2 \quad r=2$
Hence required probability $={ }^n C_r p^r \cdot q^{n-r}$
$={ }^2 C_2\left(\frac{4}{5}\right)^2 \cdot\left(\frac{1}{5}\right)^0=\frac{16}{25}$.
$P_{\text {(defected) }}=\frac{2}{10}=\frac{1}{5}=q$ and $n=2 \quad r=2$
Hence required probability $={ }^n C_r p^r \cdot q^{n-r}$
$={ }^2 C_2\left(\frac{4}{5}\right)^2 \cdot\left(\frac{1}{5}\right)^0=\frac{16}{25}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.