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Two cards are drawn with replacement from a well shuffled deck of $52$ cards. Find the mean and variance for the number of aces.

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  • 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) $\sum P_i=1$
  • Moments of a discrete random variable :
  • (i) About the origin : $\mu_r'=E(X^r)=\sum P_ix_i^{\Large r}$
  • First moment : $\mu_1'=E(X)=\sum P_ix_i$
  • Second moment : $\mu_2'=E(X^2)=\sum P_ix_i^2$
  • (ii) About the mean : $\mu_n=E(X-\bar{X})^n=\sum (x_i-\bar{x})^nP_i$
  • First moment : $\mu_1=0$
  • Second moment : $\mu_2=E(X-\bar{X})^2=E(X^2)-[E(X)]^2=\mu_2'-(\mu_1')^2$
  • $\mu_2=Var(X)$
Step 1:
Let $X$ be the random variable denoting the number of aces when 2 cards are drawn,with replacement from well-shuffled pack of 52 cards.
$X$ takes the values 0,1,2
Step 2:
$P(X=0)$=Probability of no aces
$\qquad\quad\;=\large\frac{48}{52}\times \frac{48}{52}$
$P(X=1)=$Probability of 1 ace
$\qquad\quad\;=2C_1\large\frac{4}{52}\times \frac{48}{52}$
$\qquad\quad\;=2\times\large\frac{1}{13}\times \frac{12}{13}$
$P(X=2)=$Probability of 2 aces
$\qquad\quad\;=\large\frac{4}{52}\times \frac{4}{52}$
Step 3:
The probability distribution of $X$ is given by
Step 4:
Mean =$E(X)=\sum x_iP_i$
$\qquad=0\times \large\frac{144}{169}$$+1\times \large\frac{24}{169}$$+2\times \large\frac{1}{169}$
Step 5:
$E(X^2)=\sum x_i^2P_i$
$\qquad\;\;=0\times \large\frac{144}{169}$$+1\times \large\frac{24}{169}$$+4\times \large\frac{1}{169}$
$\therefore Var(X)=\large\frac{28}{169}-\frac{4}{169}$
answered Sep 17, 2013 by sreemathi.v

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