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# 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.

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

• 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 $(q = 1 – p)$
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 $(q = 1 – p)$
(i) $P (X = 0) = \large^{5}C_0$$\times (0.05)^0 \times (0.95)^5 = 0.95^5 (ii) P (X \leq 1 ) = P(X=0) + P (X=1) P (X \leq 1) = 0.95^5+ \large^{5}C_1$$ \times (0.05)^1 \times (0.95)^(5-1) = 0.95^5+0.95^4(5\times0.05)$P (X \leq 1) = 0.95^4 (0.95+0.25) = 1.2 \times 0.95^4$(iii)$P(X\gt1) = 1 - P (X\leq1) = 1 - 1.2 \times 0.95^4$(iv)$P (X \geq 1) = 1 - P (X = 0) = 1 - 0.95^5\$