Question

The probability that a bulb, produced by a factory, will fuse after $150$ days if used is $0.05$. If the probability that out of $5$ such bulbs, none will fuse after $150$ days of use is ${\left( {\dfrac{{19}}{k}} \right)^5}$ , then what is the value of $k$ ?

Answer 1

Fusing of bulbs is a Bernoulli Trial. Bernoulli Trials are trials with only two outcomes, that is, success and failure. The probability of success and failure of a Bernoulli Trial is always the same for a trial of an event. Use the probability formula for the binomial distribution of a Bernoulli Trial to solve the above question.

Formula used: For an event, the binomial probability of an outcome of an event is given by:
$P = {}^nC{}_x{p^x}{\left( {1 - p} \right)^{n - x}}$
Where $n$ is the number of trials,
$k$ is the number of times the outcome occurs,
$p$ is the probability of the success of the outcome and
$\left( {1 - p} \right)$ is the probability of the failure of the outcome.

Complete step by step Solution:
The probability that the bulb will fuse after $150$ days if used gives the probability of the success of the trial.
Therefore, $p = 0.05$ .
The number of bulbs gives us the total number of trials.
Therefore, $n = 5$ .
Now, we are asked to consider a trial where none of the bulbs fused, therefore, $x = 0$.
Hence, the binomial probability, $P = {\left( {\dfrac{{19}}{k}} \right)^5}$ .
We know that the binomial probability of a Bernoulli Trial is given by:
$P = {}^nC{}_x{p^x}{\left( {1 - p} \right)^{n - x}}$
Substituting all the values,
${\left( {\dfrac{{19}}{k}} \right)^5} = {}^5{C_0}{\left( {0.05} \right)^0}{\left( {1 - 0.05} \right)^5}$
Simplifying further,
${\left( {\dfrac{{19}}{k}} \right)^5} = {\left( {0.95} \right)^5}$
As the exponents are the same, thus, equating the bases,
$\dfrac{{19}}{k} = 0.95$
This gives: $k = 20$
Thus, the value of $k$ is $20$.

Note: The value of Binomial Probability for different trials will be different but the probability of their success and failure will remain the same. In the above question, the value of $x$ will vary along with the conditions given in the question.