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Home  >>  CBSE XII  >>  Math  >>  Probability
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A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \( \frac{1}{100}\). What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice?

$\begin{array}{1 1} (a)\; 1 - \large(\frac{99}{100})^{50} \quad (b)\; \large\frac{1}{2} (\frac{99}{100})^{49} \quad (c)\;1 - \large(\frac{99}{100})^{49}(\frac{149}{100}) \\(a)\; 1 - \large(\frac{99}{100})^{50} \quad (b)\; \large\frac{1}{2} (\frac{49}{100})^{49} \quad (c)\;1 - \large(\frac{49}{100})^{49}(\frac{99}{100}) \\(a)\; 1 - \large(\frac{99}{50})^{25} \quad (b)\; \large\frac{1}{2} (\frac{49}{50})^{49} \quad (c)\;1 - \large(\frac{49}{50})^{49}(\frac{49}{100}) \\ (a)\; 1 - \large(\frac{99}{50})^{25} \quad (b)\; \large\frac{1}{2} (\frac{99}{100})^{49} \quad (c)\;1 - \large(\frac{49}{50})^{49}(\frac{149}{100}) \end{array} $

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1 Answer

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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.
Out of $(n=50)$ lotteries, let X denote the number of times the person wins a prize, with $p = \large\frac{1}{100}$ and $q = 1 - p = \large\frac{99}{100}$.
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.
Since 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 $(q = 1 – p)$
a) P (wins at least once) $= P(X \gt 0) = 1 - P (X = 0)$
$P (X \gt 0) = 1 - ^{50}C_0. \large\frac{1}{100}^0.\large\frac{99}{100}^{50–0}$ $=1 - \large(\frac{99}{100})^{50}$
b) P ( wins exactly once) $ = P (X = 1)$
$P (X = 1) = ^{50}C_1. \large\frac{1}{100}^1.\large\frac{99}{100}^{50–1}$ $ = 50 \large\frac{1}{100}\frac{99}{100}^{49} = \frac{1}{2} (\frac{99}{100})^{49}$
c) wins at least twice) $ = P (X \geq 2) = 1 - P ( X \leq1) = 1 - P (X =0) - P (X=1)$
$P (X \geq 2) = 1 - \large(\frac{99}{100})^{50} - \frac{1}{2} (\frac{99}{100})^{49}$$ = 1 - \large(\frac{99}{100})^{49}(\frac{99+50}{100})$
$P (X \geq 2) = 1 - \large(\frac{99}{100})^{49}(\frac{149}{100})$

 

answered Jun 21, 2013 by balaji.thirumalai
 

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