Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.

Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.    

Answers (1)

Let X be a random variable of number of aces

X can take values 0, 1 or 2 because only two cards are drawn.

Therefore, Total deck of cards = 52
and total no. of ACE cards in a deck of cards = 4

Since the draws are done without replacement, therefore, the two draws are not independent.
Therefore,
P(X = 0) = Probability of no ace being drawn
= P(non – ace and non – ace)
= P(non – ace) × P(non – ace)

\\=\frac{48}{52} \times \frac{47}{51}$ \\$=\frac{2256}{2652}$ \\$\mathrm{P}(\mathrm{X}=1)=$ Probability that 1 card is an ace \\$=$ P(ace and non - ace or non -ace and ace) \\$=$ P(ace and non $-$ ace $)+P($ non $-$ ace and $a c e)=P(a c e) P($ non $-a c e)+P($ non $-$ ace $)$ P(ace) \\$=\frac{4}{52} \times \frac{48}{51}+\frac{48}{52} \times \frac{4}{51}$ \\$=\frac{384}{2652}$ \\$=\frac{192}{2326}=\frac{96}{1613}$ \\$\mathrm{P}(\mathrm{X}=2)=$ Probability that both cards are ace \\$=$ P(ace and ace) \\$=\mathrm{P}(\mathrm{ace}) \times \mathrm{P}(\mathrm{ace})$ \\$=\frac{4}{52} \times \frac{3}{51}$

\begin{aligned} &=\frac{12}{2652}=\frac{6}{1326}=\frac{3}{663}=\frac{1}{221}\\ &\text { As we know that, }\\ &\operatorname{Mean}(\mu)=E(X)=\Sigma X P(X)\\ &=0 \times \frac{2256}{2652}+1 \times \frac{384}{2652}+2 \times \frac{12}{2652}\\ &=\frac{384}{2652}+\frac{24}{2652}\\ &=\frac{408}{2652} \end{aligned}

\\ =\frac{2}{13} \\ \text { Also, } \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2} \\ =\Sigma X^{2} P(X)-[E(X)]^{2} \\ =\left[0 \times \frac{2256}{2652}+1^{2} \times \frac{384}{2652}+2^{2} \times \frac{12}{2652}\right]-\left(\frac{2}{13}\right)^{2} \\ =\frac{432}{2652}-\frac{4}{169} \\ =0.1629-0.0237 \\ =0.1392 \\ \therefore \text { Standard Deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{ 0.1392 }\cong 0.373(\text { approx. })

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question