Question

If two cards are drawn randomly from a pack of 52 playing cards, then the mean of the probability distribution of number of kings is:

• A. \(\frac{215}{221}\)
• B. \(\frac{2}{13}\)
• C. \(\frac{188}{221}\)
• D. \(\frac{13}{2}\)

Hint:

For drawing without replacement, use combinations. The mean of a probability distribution is \mu = \sum x P(X=x).

Answer 1

The Correct Option is B

Step 1: Define the random variable. Let X be the random variable representing the number of kings drawn.
The possible values of X are 0, 1, and 2.

Step 2: Calculate the probabilities for each value of X.
Total number of ways to draw 2 cards from 52 is \(\binom{52}{2} = \frac{52 \times 51}{2} = 1326\).
 

P(X=0): No kings drawn. Number of ways to choose 2 non-king cards from 48 is \(\binom{48}{2} = \frac{48 \times 47}{2} = 1128\).
P(X=0) = \(\frac{1128}{1326} = \frac{188}{221}\)

P(X=1): One king drawn. Number of ways to choose 1 king from 4 and 1 non-king from 48 is \(\binom{4}{1} \times \binom{48}{1} = 4 \times 48 = 192\).
P(X=1) = \(\frac{192}{1326} = \frac{32}{221}\)

P(X=2): Two kings drawn. Number of ways to choose 2 kings from 4 is \(\binom{4}{2} = \frac{4 \times 3}{2} = 6\).
P(X=2) = \(\frac{6}{1326} = \frac{1}{221}\)

 

Step 3: Calculate the mean of the probability distribution.
Mean (\(\mu\)) = \(\sum x P(X=x)\)
\(\mu = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)\)
\(\mu = 0 \times \frac{188}{221} + 1 \times \frac{32}{221} + 2 \times \frac{1}{221}\)
\(\mu = 0 + \frac{32}{221} + \frac{2}{221} = \frac{34}{221} = \frac{2}{13}\)
Therefore, the mean of the probability distribution of the number of kings is \(\frac{2}{13}\).