Question

Three cards are randomly selected, without replacement, from an ordinary deck of $52$ playing cards. Compute the conditional probability that the first card selected is a spade given that the second and third cards are spades.

Short answer

$0.22$isthe conditional probability that the first card selected is a spade given that the second and third cards are spades.

Step-by-step solution

Step 1:Given Information

Three cards are arbitrarily chosen without substitution from a normal deck of $52$cards.

In a deck of cards, the quantity of spades is$13$

Allow$A$to signify the occasion that the principal card is a spade.

Allow $B$to signify the occasion that the subsequent card is a spade.

Allow $C$to signify the occasion that the third card chose is a spade.

For three occasions $A,B,$and $C$, the conditional probability $P(A\mid B\cap C)$is characterized as:

$P(A\mid B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}$

From the multiplication rule of the probability of three occasions, we know that:

$P(A\cap B\cap C)=P\left(A\right)P(B\mid A)P(C\mid A\cap B)$

From the law of complete probability rule for the three occasions, we know that:

$P(B\cap C)=P\left(A\right)P(B\mid A)P(C\mid A\cap B)+P\left(A^c\right)P\left(B\mid A^c\right)P\left(C\mid A^c\cap B\right)$

Step 2:Explanation of P(A)P(B∣A)P(C∣A∩B)

Now let us discover the probabilities

$P\left(A\right),P(B\mid A),P(C\mid A\cap B),P\left(B\right)$and $P\left(A^c\right)$$P\left(B\mid A^c\right),P\left(C\mid A^c\cap B\right)$

The probability that the first card select a spade card is,

$P\left(A\right)=\frac{\text{Number of spades}}{\text{Total number of cards}}$

$=\frac{13}{52}$

The probability that the second card is spade given that the chosen first card spade is,

$P(B\mid A)=\frac{12}{51}\left(\begin{array}{l}\text{Since one spade is already selected in the first draw, we}\\ \text{are left with only}12\text{spades and}51\text{total cards}\end{array}\right)$

The probability that the third card chosen is a spade given that the principal card is a spade and the second is a spade is.

$P(C\mid A\cap B)=\frac{11}{50}\left(\begin{array}{l}\text{Since two spades are already selected in the first}\\ \text{and second draws, we are left with only}11\text{spades}\\ \text{and}50\text{total cards}\end{array}\right)$

Step 3:Explanation of PAcPB∣AcPC∣Ac∩B

The probability that the first card is chosen, not a spade card is,

$P\left(A^c\right)=1-P\left(A\right)$

$=1-\frac{13}{52}$

$=\frac{39}{52}$

The probability that the second card chosen is a spade given that the principal card isn't a spade is,

$P\left(B\mid A^c\right)=\frac{13}{51}\left(\begin{array}{l}\text{Since first card is not a spade card that is any of non}\\ \text{spade and we have to select second card is spade of}\\ \text{all}13\text{spade cards from the total of}51\text{cards.}\end{array}\right)$

The probability that the third card chosen is a spade given that the main card is a non-spade card and the second a spade is,

$P\left(C\mid A^e\cap B\right)=\frac{12}{50}\left(\begin{array}{l}\text{Since first card is non spade card and the}\\ \text{second is a spade card and finally we have}\\ \text{to select third is card is spade of all}12\\ \text{spade cards from the total of}50\text{cards.}\end{array}\right)$

Step 4:Final Answer

The conditional probability that the primary card chose is a spade allowed that the second and third cards are spades is given by:$P(A\mid B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}$

$=\frac{P\left(A\right)P(B\mid A)P(C\mid A\cap B)}{P\left(A\right)P(B\mid A)P(C\mid A\cap B)+P\left(A^c\right)P\left(B\mid A^c\right)P\left(C\mid A^c\cap B\right)}$

$=\frac{\frac{13}{52}\times \frac{12}{51}\times \frac{11}{50}}{\frac{13}{52}\times \frac{12}{51}\times \frac{11}{50}+\frac{39}{52}\times \frac{13}{51}\times \frac{12}{50}}$

$=\frac{0.0129}{0.0129+0.0459}$

$=\frac{0.0129}{0.0588}$

$=0.22$

In this way, the conditional probability that the primary card chosen is a spade allowed that the second and third cards are spades is $0.22$.