Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings.

 

 

 

 
 
 
 
 

Answers (1)

Let S = no. of Kings

and F = number of non-kings

        \therefore P(S) = \frac{4}{52} and P(F) = \frac{48}{52}

Let x_i be the event that the king is drawn

x_i

Px_i x_iPx_i x_i^2 Px_i \cdot x_i^2
0 \frac{48}{52}\times \frac{47}{51}\times 1 = \frac{2256}{2652} 0 0 0
1 \frac{4}{52}\times \frac{48}{51}\times 2 = \frac{384}{2652} \frac{384}{2652} 1

\frac{384}{2652}

2 \frac{4}{52}\times \frac{3}{51}\times 1 = \frac{12}{2652} \frac{12}{2652} 4

\frac{48}{2652}

 

Px_i

\frac{2256}{2652} \frac{384}{2652}

\frac{12}{2652}

x_i

0 1 2

Mean    \overline{x} = \sum x_i^2P(x_i)

              \overline{x} =\frac{384 + 12}{2652} = \frac{396}{2652}

Variance \sigma^2= \sum x_i^2P(x_i)-(\overline{x})^2

                        = \frac{432}{2652} -\frac{396}{2652}^2

                        = \frac{36}{221} -\frac{33}{221}^2

                        \Rightarrow \frac{(36\times 221)-(33\times 33)}{221\times 221}

                        \Rightarrow \frac{7956 - 1089}{48841} = \underline{\frac{6867}{48841}}

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