Q. 2.19 - A First Course in Probability

Question

Ten cards are randomly chosen from a deck of $52$cards that consists of$13$ cards of each different suit. Each of the selected cards is put in one$4$ pile, depending on the suit of the card.

$\left(a\right)$What is the probability that the largest pile has $4$cards, the next largest has$3$, the next largest has$2$, and the smallest has $1$cards?

$\left(b\right)$What is the probability that two of the piles have $3$cards, one has $4$cards, and one has no cards?

Step-by-Step Solution

Answer

$a)$$P\left(\right(4,3,2,1)$cards in each suit$)\approx 31.457\%$

$b)$ $P\left(\right(4,3,3,0)$cards in each suit$)\approx 4.436\%$

Step 1 Given Information.

Ten cards are randomly chosen from a deck of $52$cards that consists of$13$ cards of each $4$different suit. Each of the selected cards is put in one $4$pile, depending on the suit of the card.

Step 2 Explanation.

Outcome space $S$contains every combination (a subset of fixed size) of cards. If all events  $S$ are considered equally likely, the probability of an event $A\subseteq S$ is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where$\left|X\right|$ denotes the number of elements in $X$the chapter$1.4$. it is shown that the number of ten card combinations of $52$different cards is$\left(\begin{array}{l}52\\ 10\end{array}\right)=\left|S\right|$.

Step 3 part (a) Explanation.

$A$is the event that the numbers of cards from the four suits are$4,3,2,1, P\left(A\right)=?$

Count all possible events from $S$in$A$.

The outcomes in $A$differentiating firstly in which suit has how many cards. Choose the suit from which $A$cards are drawn in 4 ways, the one with cards drawn from the remaining$3$ suits, etc. so $4 !$ways to determine that.

The number of ways in which we can pick the $4$cards from the chosen suit is$\left(\begin{array}{c}13\\ 4\end{array}\right)$ - combinations of $4$out of $13$cards of that suit. Three cards from another suit in $\left(\begin{array}{c}13\\ 3\end{array}\right)$ways, two cards from the thirteen in $\left(\begin{array}{c}13\\ 2\end{array}\right)$ways, and the one card from the suit from which the leas cards were drawn in $\left(\begin{array}{c}13\\ 1\end{array}\right)$ways.

Now only apply the product principle

$P\left(\right(4,3,2,1\left)\text{ cards in each suit }\right)$$=\frac{4!\left(\begin{array}{c}13\\ 4\end{array}\right)\left(\begin{array}{c}13\\ 3\end{array}\right)\left(\begin{array}{c}13\\ 2\end{array}\right)\left(\begin{array}{c}13\\ 1\end{array}\right)}{\left(\begin{array}{c}52\\ 10\end{array}\right)}\cong 31457$

Step 4 Part (b) Explanation.

$B$is the event that the numbers of cards from the four suits are$4,3,3,0, P\left(B\right)=$?

Count all possible events from $S$to$B$.

The outcomes $B$differentiate firstly in which suit has how many cards. Choose the suit from which $4$cards are drawn in $4$ways, the one with $0$cards drawn from the remaining$3$ suits, and the suits from which the three cards are drawn are shouldn't be permuted because they are no different. Total of $4·3$different assignments of suits to the numbers of cards.

The number of ways in which we can pick the $4$cards from the chosen suit is$\left(\begin{array}{c}13\\ 4\end{array}\right)$- combinations of $4$out of$13$ cards of that suit. Three cards from the determined suit in $\left(\begin{array}{c}13\\ 3\end{array}\right)$ways and the remaining cards from the third chosen suit in \|binom $133$ways.

Now only apply the product principle

$P\left(\right(4,3,3,0\left)\text{ cards in each suit }\right)$$=\frac{4.3.\left(\begin{array}{c}13\\ 4\end{array}\right)\left(\begin{array}{c}13\\ 3\end{array}\right)\left(\begin{array}{c}13\\ 3\end{array}\right)}{\left(\begin{array}{c}52\\ 10\end{array}\right)}\cong 0.04436$