The face cards are removed from a full pack. Out of the remaining 40 cards, 4 are drawn at random. what is the probability that they belong to different suits?
Solution
Having removed 12 face cards, the remaining 40 cards include 10 cards in each suit.
∴ Chance of drawing a card in the first draw = \[\frac{^{40}{}{C}_1}{^{40}{}{C}_1} = 1\]
Having drawn 1 card, there remain 39 cards of which 30 are of suits different from the drawn card.
∴ Chance of drawing a card of different suit in the second draw = \[\frac{^{30}{}{C}_1}{^{39}{}{C}_1} = \frac{30}{39}\]
Having drawn two cards, there remain 38 cards of which 20 are of suits different from the drawn cards.
∴ Chance of drawing a card of in the third draw =\[\frac{^{20}{}{C}_1}{^{38}{}{C}_1} = \frac{20}{38} = \frac{10}{19}\]
Having drawn three cards, there remain 37 cards of which 10 are of suits different from the drawn cards.
∴ Chance of drawing a card in the fourth draw =\[\frac{^{10}{}{C}_1}{^{37}{}{C}_1} = \frac{10}{37}\]
Hence, all the events being dependent, the required probability =\[1 \times \frac{30}{39} \times \frac{10}{19} \times \frac{10}{37} = \frac{1000}{9139}\]
