Question

The probability that a bulb produced by a factory will fuse after 180 days of use is 0.05. The probability that out of 5 such bulbs none fuse after 180 days is \[{{\left( \dfrac{19}{20} \right)}^{k}}\]. The value of ‘k’ is _____.

Answer 1

Hint:Calculate the probability of one bulb to not fuse after 180 days by equation, \[P\left( E \right)+P\left( E' \right)=1\], where P(E)is the probability of occurring of any event and \[P\left( E' \right)\]is the probability of not occurring of it. Now, calculate the probability of five bulbs such as none of them will fuse after 180 days by multiplying the probabilities of each bulb to not fuse individually.

Complete step by step answer:
It is given that a bulb will be fused after 180 days with the probability of 0.05. And we need to determine the value of k, if the probability of 5 such bulbs and none will fuse after 180 days is \[{{\left( \dfrac{19}{20} \right)}^{k}}\].
Now, we know that sum of probability of occurring of any event and not occurring of any event and not occurring of it is 1 i.e. \[P\left( E \right)+P\left( E' \right)=1\]. Hence, we can calculate the probability of not fusing a bulb after 180 days by relation.
\[P\left( E \right)+P\left( E' \right)=1\]- (i)
Where P (E) denotes probability of fusing the bulb and \[P\left( E' \right)\]is the probability of not fusing the bulb. So, we know that a bulb will fuse with probability 0.05 after 180 days. Hence, we have
P (E) = 0.05
\[\Rightarrow P\left( E \right)=\dfrac{5}{100}=\dfrac{1}{20}.........(ii)\]
Now, we can calculate probability of not fusing a bulb by relation (i), we get
\[P\left( E \right)+P\left( E' \right)=1\]
\[\begin{align}
  & \dfrac{1}{20}+P\left( E' \right)=1 \\
 & P\left( E' \right)=1-\dfrac{1}{20}=\dfrac{20-1}{20} \\
 & P\left( E' \right)=\dfrac{19}{20}-(iii) \\
\end{align}\]
Hence, the probability of not fusing a bulb after 180 days is \[\dfrac{19}{20}\].
So, we can calculate the probability for 5 bulbs, none of them will fuse after 180 days. Hence, it can be calculated as
\[P=P\left( E_{1}^{'} \right)\times P\left( E_{2}^{'} \right)\times P\left( E_{3}^{'} \right)\times P\left( E_{4}^{'} \right)\times P\left( E_{5}^{'} \right)\]
Where \[P\left( E_{1}^{'} \right),P\left( E_{2}^{'} \right),P\left( E_{3}^{'} \right),P\left( E_{4}^{'} \right)\]and \[P\left( E_{5}^{'} \right)\]are the individual probabilities of not fusing after 180 days.
Hence, we can put values of \[P\left( E_{1}^{'} \right),P\left( E_{2}^{'} \right),P\left( E_{3}^{'} \right),P\left( E_{4}^{'} \right),P\left( E_{5}^{'} \right)\]as \[\dfrac{19}{20}\]by the equation (iii) as probability of not diffusing for a single bulb will be same for each. Hence, we get
\[\begin{align}
  & P={{\left( P\left( E' \right) \right)}^{5}} \\
 & P={{\left( \dfrac{19}{20} \right)}^{5}} \\
\end{align}\]
Now, we can compare the given probability with the above calculated probability. So, we have probability that out of 5 bulbs none fuse after 180 days is \[{{\left( \dfrac{19}{20} \right)}^{k}}\], from the question.
Hence, by comparing both, we get k = 5.

Note: Another approach for the question would be that we can write the probability of given event by binomial theorem of probability given as
\[P={}^{n}{{C}_{r}}{{P}^{r}}{{q}^{n-r}}\]where, q = 1-p.
So, we need to calculate the probability that out of 5 bulbs none fuse after 180 days.
So, the probability bulb will fuse = 0.05.
So, none of 5 will fuse can be given as
\[\begin{align}
  & P={}^{5}{{C}_{0}}{{\left( 0.05 \right)}^{0}}{{\left( 1-0.05 \right)}^{5}} \\
 & P={{\left( 0.95 \right)}^{5}}={{\left( \dfrac{19}{20} \right)}^{5}} \\
\end{align}\]
One may add the probabilities \[P\left( E_{1}^{'} \right),P\left( E_{2}^{'} \right),P\left( E_{3}^{'} \right),P\left( E_{4}^{'} \right),P\left( E_{5}^{'} \right)\]in the solution which is wrong. We want to determine the probability of 5 bulbs combindly, so we can’t add them. So be clear with the fundamental properties of multiplication and addition of probability.