Answer to Question #206372 in Statistics and Probability for Chilu

Question

I. Two cards are drawn from a well-shuffled ordinary deck of 52 cards. Find the probability that they are both aces if the first card is  a) replaced,  b) not replaced. II. A ball is drawn at random from a box containing 6 red balls, 4 white balls, and 5 blue balls. Determine the probability that it is (a) red, (b) white, (c) blue, (d) not red, (e) red or white. III. Prove that the mean and variance of a binomially distributed random variable are, respectively, µ = np and σ2 = npq. 

Answer 1

(1). Probability of drawing two aces from 52 cards is-

(a) If first is replaced-

$P=\dfrac{4}{52}\times \dfrac{4}{52}=\dfrac{1}{169}$

(b) If first is not replaced-

$p=\dfrac{4}{52}\times \dfrac{3}{51}=\dfrac{1}{221}$

(2). Given, 6 Red balls, 4 White balls, 5 blue balls-

Total balls=15

$(a) P(\text{Red ball})=\dfrac{6}{15}=\dfrac{2}{5} \\[9pt] (b) P(\text{White ball})=\dfrac{4}{15}\\[9pt] (c) P(\text{Blue ball})=\dfrac{5}{15}=\dfrac{1}{3}\\[9pt] (d) P(\text{not a Red ball})=1-P(Red ball)=1-\dfrac{2}{5}=\dfrac{3}{5} \\[9pt] (e) P( \text{Red or White})= P(R)+P(W)=\dfrac{6}{15}+\dfrac{4}{15}=\dfrac{10}{15}=\dfrac{2}{3}$

(3).

Mean=$Ex=\sum x \text{ }^nC_x p^xq^{(1-x)}=np$

$x\times \dfrac{n!}{(n-x)! x!}=x\dfrac{n(n-1)!}{(n-x)! x (x-1)!}=n\text{ }^{n-1} C_{x-1}$

So Mean $= \sum n\text{ }^{n-1}C_{x-1}p^{x-1}.p.(1-p)^{n-x}$

   $np\sum ^{n-1}C_{x-1}$

Let x-1=y, n-1=m

So Mean=$np \sum ^mC_y P^y(1-p)^{m-y}$

As we know $\sum f(x)=1$

So Mean$=np\times 1=np$

Variance $= E[x^2]-[E(x)]^2=\sum x(x-1) \text{ }^n C_xp^xq^{(1-x)}$

      

So $E[x(x-1)]=\sum x(x-1) \text{ } ^nC_xp^xq^{n-x}$

       

  $x(x-1) \text{ } ^nC_x=x(x-1)\dfrac{n!}{(n-x)! x!}=x(x-1)\dfrac{n(n-1)(n-2)!}{(n-x)!x(x-1)(x-2)!}$

       

$=n(n-1)\dfrac{(n-2)!}{(n-x)!(x-2)!} \\[9pt] =n(n-1)\text{ } ^{n-2}C_{x-2}$

So $E[x(x-1)]=\sum n(n-1) \text{ } ^{n-2}C_{x-2} p^{x-2} (1-p)^{n-x}$

      

$=(n(n-1))p^2 \sum \text{ } ^{n-2}C_{x-2} p^{x-2}(1-p)^{n-x}$

           Let x-2=y,n-2=m

     $=n(n-1)p^2 \sum \text{ } ^mC_y p^y(1-p)^{m-y}$

      $=n(1-n)p^2 \times 1$   

As $\sum f(x)=1$

Variance=$E(X^2)-[E(x)]^2$

     $=E[x(x-1)]+E(X)-[E(x)]^2 \\ =n(n-1)p^2+np-(np)^2 \\ =n^2p^2-np^2+np-n^2p^2 \\ =np(1-p)$