Question

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of such 5 bulbs.
I.None
II.Not more than one
III.More than one
IV.At least one will fuse after 150 days

Answer 1

Hint : In such type of questions in which only two cases are possible first of success of any event and second is failure of the event, we can use Binomial distribution. According to binomial distribution, if n be the number of observation or event, P be the probability of success (success is not only positive outcome it could also be negative for example if in any question probability of an accident is given so we will consider (accident as success) and q be the probability of failure and r be the unique order of success and $\left( {n - r} \right)$ order of failure.
Then,
$P(X = r){ = ^h}{C_r}{P^r}{q^{n - r}}$
Where $r{\text{ }} = {\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }} \ldots \ldots n$ and $p + q = 1$

Complete step-by-step answer :
Given: the time for the bulb to get fused is 150 days and the probability is 0.5.
According to question,
Probability that bulb will fuse after 150 days (Probability of success), $P{\text{ }} = {\text{ }}0.05$
So, probability that bulb will not fuse after 150 days is $q{\text{ }} = {\text{ }}1{\text{ }}-{\text{ }}0.05{\text{ }} = {\text{ }}0.95$
The number of independent events = 5
i.None bulb will fuse, means $r{\text{ }} = {\text{ }}0$
So,
$P(X = 0){ = ^5}{C_0}{(0.05)^0} \times {(0.95)^{5 - 0}}$
$P(X = 0) = 1 \times 1 \times {(0.95)^5}$
$P(X = 0) = {(0.95)^5}$
ii.Not more than on bulb will fuse will fuse means $r{\text{ }} \leqslant {\text{ }}1$
So required probability,
$P(X \leqslant 1) = P(X = 0) + P(X = 1)$
$P(X \leqslant 1) = {(0.95)^5}{ \times ^5}{C_1} \times {(0.05)^1} \times {(0.95)^{5 - 1}}$
$P(X \leqslant 1) = {(0.95)^5} \times 5 \times (0.05) \times {(0.95)^4}$
$P(X \leqslant 1) = {(0.95)^4}(0.95 + 0.25)$
$P(X \leqslant 1) = 1.2 \times {(0.95)^4}$

iii.More than one bulb will fuse means $r > 1$
So required probability,
$P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$ ……….. (i)
Since total probability = 1
So, $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1$
$\Rightarrow P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1 - [P(X = 0) + P(X = 1)]$
$\Rightarrow 1 - P(X \leqslant 1)$
So, we can also write equation (i) as
$P(X > 1) = 1 - P(X \leqslant 1)$
$P(X > 1) = 1 - 1.2 \times {(0.95)^4}$
iv.At least one bulb will fuse means $r{\text{ }} \geqslant {\text{ }}1$
$P(X \geqslant 1) = 1 - P(X = 0)$
$\Rightarrow 1 - {(0.95)^5}$

Note : You may confuse that in which question binomial distribution is used. Here are the criteria to use binomial distribution.
a.The number of observations (n) is fixed.
b.Each observation is independent in other words none of your trials have an effect on other trials.
c.The probability of success is the same for each trail.