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Home  >>  CBSE XII  >>  Math  >>  Probability
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Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that (i) all the five cards are spades? (ii) only 3 cards are spades? (iii) none is a spade?

$\begin{array}{1 1}(i) \large\frac{1}{1024}\quad(ii) \large\frac{90}{1024}\quad(iii) \large\frac{243}{1024}\quad \\ (i) \large\frac{10}{1024}\quad(ii) \large\frac{9}{1024}\quad (iii) \large\frac{23}{1024}\quad \\(i) \large\frac{10}{1024}\quad (ii) \large\frac{45}{1024}\quad(iii) \large\frac{243}{1024}\quad \\(i) \large\frac{1}{1024}\quad(ii) \large\frac{45}{1024}\quad(iii) \large\frac{23}{1024}\quad\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.
There are 52 cards in a deck, out of which 13 are spades.
The experiment is to draw (n = 5) cards w replacement. 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 ( a spade is drawn ) = p = $\large\frac{13}{52} = \frac{1}{4}$
P ( a card other than spade is drawn ) = q = 1 - p = $1 - \large\frac{13}{52} = \frac{3}{4}$
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 $(q = 1 – p)$
Here $n=5, p = \large\frac{1}{4}, \;$$q=\large\frac{3}{4}$.
(i) Probability that if 5 cards are drawn all of them are spades:
$P (X = 5) = \large^{5}C_5. \large\frac{1}{4}^5.\frac{3}{4}^{5–5}=$ $\large\frac{1}{4}^5 = \frac{1}{1024}$
(ii) Probability that only 3 cards are spades = P (3 spades and 2 non spades):
P (X = 3) = $\large^{5}C_3\large\frac{1}{4}^3$$\times$$\large\frac{3}{4}^5$ = $\large\frac{60}{6}\frac{1}{64}$$\times$$\large\frac{9}{16} = \frac{90}{1024}$
(iii) Probability that there are no spades = P (0 spades):
P (X = 0) = (ii) Probability that only 3 cards are spades = P (3 spades and 2 non spades):
$P (X = 0) = \large^{5}C_0. \large\frac{1}{4}^0.\frac{3}{4}^{5–0}=(\frac{3}{4})^5 = \frac{243}{1024}$
answered Jun 21, 2013 by balaji.thirumalai
 

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