Question

Two cards are drawn at random from a shuffled deck and laid aside without beingexamined. Then a third card is drawn. Show that the probability that the thirdcard is a spade is ¼ just as it was for the first card. Hint: Consider all the (mutuallyexclusive) possibilities (two discarded cards spades, third card spade or not spade,etc.).

Short answer

Answer

The probability that the third card is a spade is$\frac{1}{4}$.

Step-by-step solution

Given Information

Two cards are drawn and kept aside without looking and a third card is drawn

Definition of Independent Event

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

When events are independent, apply the formula $P\left(AB\right).P\left(B\right)$where A and B are the events.

Important Information of event

In a deck of card, there are 52 cards of 4 suits namely spade, club, diamond and heart, out of which spade and club are black and diamond and heart are red.

There are 13 cards of each suit and 26 cards of each color.

Proving the statement

When the first card is drawn, the probability of getting a spade is $\frac{13}{52}$and getting a non-spade is$\frac{39}{52}$.

When the second card is drawn, the probability of getting a spade when the first card was spade is $\frac{12}{51}$and a non-spade is $\frac{38}{51}$. And the probability of getting a spade when the second card was non-spade is $\frac{13}{51}$and a non-spade is $\frac{38}{51}$.

Similarly, the probability of the third spade can be found.

Find the probability of the third spade.

$$\begin{gathered} \frac{12}{52}\times \frac{12}{51}\times \frac{11}{50}\times \frac{13}{52}\times \frac{38}{51}\times \frac{12}{50}+\frac{39}{52}\times \frac{39}{52}\times \frac{13}{51}\times \frac{12}{50}+\frac{39}{52}\times \frac{38}{51}\times \frac{13}{50}=\frac{32994}{132600} \\ \approx \frac{1}{4} \end{gathered}$$

The required probability is $\frac{1}{4}$. Hence the given statement is valid.