Question

10 selecting cards the red face cards and the black cards numbered 2 - 9 are put into a bag. four cards are drawn at random without replacement. find the following probabilities.
a. all 4 cards are red.
b. 2 cards are red and 2 cards are black.
c. at least 1 of the cards is red.
d. all 4 cards are black.

Explanation:

Step1: Calculate the total number of cards

There are \(6\) red face cards (\(J,Q,K\) of hearts and diamonds) and \(8\) black cards numbered \(2 - 9\) ( \(2 - 9\) of spades and clubs). So the total number of cards \(n=6 + 8=14\)
The number of ways to draw \(4\) cards out of \(14\) is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 14\) and \(r=4\). So \(C(14,4)=\frac{14!}{4!(14 - 4)!}=\frac{14\times13\times12\times11}{4\times3\times2\times1}=1001\)

Step2: Solve part (a)

The number of red cards is \(6\). The number of ways to draw \(4\) red cards out of \(6\) is \(C(6,4)=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15\)
The probability \(P=\frac{C(6,4)}{C(14,4)}=\frac{15}{1001}\)

Step3: Solve part (b)

The number of ways to draw \(2\) red cards out of \(6\) is \(C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\)
The number of ways to draw \(2\) black cards out of \(8\) is \(C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28\)
By the multiplication principle, the number of ways to draw \(2\) red and \(2\) black cards is \(C(6,2)\times C(8,2)=15\times28 = 420\)
The probability \(P=\frac{C(6,2)\times C(8,2)}{C(14,4)}=\frac{420}{1001}=\frac{60}{143}\)

Step4: Solve part (c)

The complement of the event "at least \(1\) red card" is the event "all \(4\) cards are black"
The number of ways to draw \(4\) black cards out of \(8\) is \(C(8,4)=\frac{8!}{4!(8 - 4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\)
The probability that all \(4\) cards are black is \(P(\text{all black})=\frac{C(8,4)}{C(14,4)}=\frac{70}{1001}=\frac{10}{143}\)
Using the formula \(P(A)=1 - P(\text{complement of }A)\), the probability that at least \(1\) card is red is \(P = 1-\frac{C(8,4)}{C(14,4)}=1-\frac{70}{1001}=\frac{931}{1001}=\frac{133}{143}\)

Step5: Solve part (d)

The number of ways to draw \(4\) black cards out of \(8\) is \(C(8,4)=\frac{8!}{4!(8 - 4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\)
The probability \(P=\frac{C(8,4)}{C(14,4)}=\frac{70}{1001}=\frac{10}{143}\)

Answer:

a. \(\frac{15}{1001}\)
b. \(\frac{60}{143}\)
c. \(\frac{133}{143}\)
d. \(\frac{10}{143}\)