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Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of queens.

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Toolbox:
  • If $S$ is a sample space with a probability measure and $X$ is a real valued function defined over the elements of $S$, then $X$ is called a random variable.
  • Types of Random variables :
  • (1) Discrete Random variable (2) Continuous Random variable
  • Discrete Random Variable :If a random variable takes only a finite or a countable number of values, it is called a discrete random variable.
  • Continuous Random Variable :A Random Variable X is said to be continuous if it can take all possible values between certain given limits. i.e., X is said to be continuous if its values cannot be put in 1 − 1 correspondence with N, the set of Natural numbers.
  • The probability mass function (a discrete probability function) p(x) is a function that satisfies the following properties :
  • (1) $P(X=x)=p(x)=p_x$
  • (2) $P(x)\geq 0$ for all real $x$
  • (3) $\sum p_i=1$
Step 1:
The number of queens in the pack of cards =4
Let $X$ be the discrete RV denoting the number of queens when two cards are drawn with out replacement .$X$ takes the values $0,1,2$
Step 2:
$P(X=0)$=probability of no queen being drawn
$\qquad\quad\;=\large\frac{48C_2}{52C_2}$
$\qquad\quad\;=\large\frac{48\times 47/1\times 2}{52\times 51/1\times 2}$
$\qquad\quad\;=\large\frac{564}{663}$
$P(X=1)$=probability that 1 card is a queen
$\qquad\quad\;=\large\frac{48C_1\times 4C_1}{52C_2}$
$\qquad\quad\;=\large\frac{48\times 4}{52\times 51/1\times 2}$
$\qquad\quad\;=\large\frac{96}{663}$
$\qquad\quad\;=\large\frac{32}{221}$
$P(X=2)$=probability that both cards are queens
$\qquad\quad\;=\large\frac{4C_2}{52C_2}$
$\qquad\quad\;=\large\frac{4\times 3/1\times 2}{52\times 51/1\times 2}$
$\qquad\quad\;=\large\frac{3}{663}$
$\qquad\quad\;=\large\frac{1}{221}$
Step 3:
The probability distribution is given by
answered Sep 16, 2013 by sreemathi.v
 

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