Example: Drawing Two Cards from a Well-Shuffled Pack of Cards

Find the probability of the following scenarios:

• One red and one black card.
• Both cards of the same suit.
• One jack and one king.
• One red card and one card of a club.

Solution:

• Total ways to draw two cards:
  \[
  \binom{52}{2} = \frac{52 \times 51}{2} = 1326
  \]
• One red and one black card:
  \[
  \binom{26}{1} \times \binom{26}{1} = 26 \times 26 = 676 \quad \Rightarrow \quad P = \frac{676}{1326}
  \]
• Both cards of the same suit:
  \[
  4 \times \binom{13}{2} = 4 \times \frac{13 \times 12}{2} = 312 \quad \Rightarrow \quad P = \frac{312}{1326}
  \]
• One jack and one king:
  \[
  \binom{4}{1} \times \binom{4}{1} = 4 \times 4 = 16 \quad \Rightarrow \quad P = \frac{16}{1326}
  \]
• One red card and one card of a club:
  \[
  \binom{26}{1} \times \binom{13}{1} = 26 \times 13 = 338 \quad \Rightarrow \quad P = \frac{338}{1326}
  \]

Example: Arranging Letters in STATISTICS

If the letters of the word STATISTICS are arranged randomly, find the probability that all three T’s are together.

Solution:

• Total letters in STATISTICS: 10 (with 3 T’s, 3 S’s, 2 I’s, 1 A, 1 C).
• Total permutations:
  \[
  \frac{10!}{3! \cdot 3! \cdot 2!}
  \]
• Treating the three T’s as a single unit, total letters become 8:
  \[
  \frac{8!}{3! \cdot 2!}
  \]
• Probability:
  \[
  P = \frac{\frac{8!}{3! \cdot 2!}}{\frac{10!}{3! \cdot 3! \cdot 2!}} = \frac{8! \cdot 3!}{10!} = \frac{2}{5}
  \]

Example: Lottery Winning Probability

In a lottery, one has to choose six numbers at random out of numbers from 1 to 30. Find the probability of winning if all six chosen numbers must match the committee’s selection.

Solution:

• Total ways to choose six numbers:
  \[
  \binom{30}{6} = \frac{30 \times 29 \times 28 \times 27 \times 26 \times 25}{6!} = 593775
  \]
• Favourable outcomes (only one way to win):
  \[
  1
  \]
• Probability:
  \[
  P = \frac{1}{593775}
  \]