Question

The probability that a bulb produced by a factory will fuse after $100$ days of use is $0.05$. The probability that out of $5$ such bulbs not more than one will fuse after $100$ days is

• A. ${}^5{C}_1{\left(\frac{19}{20}\right)}^4{\left(\frac{1}{20}\right)}^1$
• B. ${\left(\frac{19}{20}\right)}^5{+}^5{C}_1{\left(\frac{19}{20}\right)}^4{\left(\frac{1}{20}\right)}^1$
• C. $1-{\left(\frac{19}{20}\right)}^5$
• D. $1-{\left(\frac{19}{20}\right)}^5{-}^5{C}_1{\left(\frac{19}{20}\right)}^4{\left(\frac{1}{20}\right)}^1$

Answer 1

The correct option is C $1-{\left(\frac{19}{20}\right)}^5$

Probability that bulb will fuse after 100 days $=0.5=\frac{1}{20}$
Probability that bulb will not fuse after 100 days $=1-\frac{1}{20}=\frac{19}{20}$
Probability that out of 5 such bulbs at least one will fuse $=1-$Probability that out of 5 such bulbs none of them fuse
$1{-}^5{C}_0\times (\frac{1}{20}{)}^0\times (\frac{19}{20}{)}^5=1-(\frac{19}{20}{)}^5$