The probability that a bulb produced by a factory will fuse after 6 months of use is $0.05 .$ find the probability that out of 5 such bulbs(i) none will fuse after 6 months of use(ii) at least one will fuse after 6 months of use

Home » The probability that a bulb produced by a factory will fuse after 6 months of use is find the probability that out of 5 such bulbs(i) none will fuse after 6 months of use(ii) at least one will fuse after 6 months of use

The probability that a bulb produced by a factory will fuse after 6 months of use is $0.05 .$ find the probability that out of 5 such bulbs
(i) none will fuse after 6 months of use
(ii) at least one will fuse after 6 months of use

The probability that a bulb produced by a factory will fuse after 6 months of use is $0.05 .$ find the probability that out of 5 such bulbs
(i) none will fuse after 6 months of use
(ii) at least one will fuse after 6 months of use

(i) The probability that the bulb will fuse $=0.05=\mathrm{p}$
The probability that the bulb will not fuse $=1-0.05=0.95=q$
Using Bernoulli’s we have,
$\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots . . \text { and } q=(1-p), n=5 \end{array}$
Probability that none will fuse $=$
$\begin{array}{l} { }^{5} C_{0} \cdot(0.05)^{0}(0.95)^{5} \\ \Rightarrow(0.95)^{5} \end{array}$
(ii) The probability that the bulb will fuse $=0.05=p$
The probability that the bulb will not fuse $=1-0.05=0.95=q$
Using Bernoulli’s we have,
$\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots . . n \text { and } q=(1-p), n=5 \end{array}$
Probability that at least one will fuse $=\mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(4)+\mathrm{P}(5)$
$\begin{array}{l} { }^{5} C_{1} \cdot(0.05)^{1}(0.95)^{4}+{ }^{5} C_{2} \cdot(0.05)^{2}(0.95)^{3}+{ }^{5} C_{3} \cdot(0.05)^{3}(0.95)^{2}+{ }^{5} C_{4} \cdot(0.05)^{4}(0.95)^{1}+{ }^{5} C_{5} \cdot(0.05)^{5}(0.95)^{0} \\ \Rightarrow\left(1-(0.95)^{5}\right) \end{array}$