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Home  >>  CBSE XII  >>  Math  >>  Probability
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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} $

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Toolbox:
  • 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
(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) $P (X \geq 1) = 1 - P (X = 0) = 1 - 0.95^5$
answered Jun 21, 2013 by balaji.thirumalai
 

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