91影视

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 $P(A鈭�B鈭�C)$is characterized as:

$P(A鈭�B鈭�C)=\frac{P(A鈭�B鈭�C)}{P(B鈭�C)}$

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

$P(A鈭�B鈭�C)=P\left(A\right)P(B鈭�A)P(C鈭�A鈭�B)$

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

$P(B鈭�C)=P\left(A\right)P(B鈭�A)P(C鈭�A鈭�B)+P\left(A^c\right)P\left(B鈭�A^c\right)P\left(C鈭�A^c鈭�B\right)$

Now let us discover the probabilities

$P\left(A\right),P(B鈭�A),P(C鈭�A鈭�B),P\left(B\right)$and $P\left(A^c\right)$$P\left(B鈭�A^c\right),P\left(C鈭�A^c鈭�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,

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

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.

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

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

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

$=1鈭�\frac{13}{52}$

$=\frac{39}{52}$

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

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,

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

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

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

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

$=\frac{0.0129}{0.0588}$

$=0.22$

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