Question

The probability that a bulb produced by a factory will fuse after $150$ days of use is $\mathrm{0.05.}$ Then the probability that out of $5$ such bulbs at least one will fuse after $150$ days of use is:

• A. $(0.95{)}^5$
• B. $(0.05{)}^5$
• C. $1-(0.95{)}^5$
• D. ${\left(\frac{1}{2}\right)}^5$

Answer 1

The correct option is C $1-(0.95{)}^5$

Let us assume that the number of bulbs that will fuse after $150$ days of use in an experiment of $5$ trials be $X$
As we can see that the trial is made with replacement, thus, the trials will be Bernoulli trials.
It is already mentioned in the question that, $p=0.05$
Thus , $q=1-p=1-0.05=0.95$
Here, we can clearly observe that $X$ has a binomial representation with $n=5$ and $p=0.05$
Thus, $P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,$ where $x=0,1,2,...n$
${=}^5{C}_x\left(0.95{)}^{5-x}\right(0.05{)}^x$
Probability of at least one such bulb will fuse in a random drawing of $5$ bulbs $=P(X\ge 1)=1-P(X<1)=1-P(X=0)=1-(0.95{)}^5$