# 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 5 such bulbs $\text{ (i) none (ii) not more than one (iii) more than one (iv) at least one}$ will fuse after 150 days of use.

$\begin{array}{1 1} (i) 0.95^5, \quad(ii) 1.2\; (0.95)^4, \quad(iii) 1 - 1.2\; (0.95)^4, \quad(iv) 1 - 0.95^5 \\ (i) 0.95^4, \quad(ii) 1.2\; (0.05)^5, \quad(iii) 1 - 1.2\; (0.05)^5, \quad(iv) 1 - 0.95^4 \\ (i) 0.95^4, \quad(ii) 1.2\; (0.95)^3, \quad(iii) 1 -\; (0.95)^4, \quad(iv) 1 - 0.95^5 \\ (i) 0.05^5, \quad(ii) 1.2\; (0.05)^4, \quad(iii) 1 - 1.2\; (0.05)^4, \quad(iv) 1 - 0.05^5\end{array}$

## 1 Answer

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• For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
• Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.
The experiment to to check if 5 bulbs drawn at random from a sample will fuse or not. It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.
P (that a bulb will fuse after 150 days) = p = 0.05.
P (that a bulb won't fuse after 150 days) = q = 1- p = 0.95
If X is the number of bulbs that will fuse after 150 days in 5 trials, X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
Here n = 5, p = 0.05 and q = 0.95

1 answer