Question

Red queens and black jacks are removed from a pack of $52$ playing cards. A cards is drawn at random from the remaining cards, after reshuffling them. Find the probability that the card drawn is
(i) a king (ii) of red colour (iii) a face card (iv) a queen

Answer 1

Total no. of cards in pack $=52$

After removing redqueens(2), and black jacks(2)

No. of cards $=52-4=48$

Solution(i):

No. of kings $=4$

Therefore, ${}^4{C}_1$( Selecting $1$ out of $4$ items) times out of ${}^{48}{C}_1$( Selecting $1$ out of $48$ items) a king is picked.

Let $E$ be the event of getting a king from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^4{C}_1}{{}^{48}{C}_1}$$=\frac{4}{48}$$=\frac{1}{12}$

Solution(ii):

No. of red colors $=26-2=24$......(2 red queen cards already drawn)

Therefore, ${}^{24}{C}_1$( Selecting $1$ out of $24$ items) times out of ${}^{48}{C}_1$( Selecting $1$ out of $48$ items) a red is picked.

Let $E$ be the event of getting red from the pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{24}{C}_1}{{}^{48}{C}_1}$$=\frac{24}{48}$$=\frac{1}{2}$

Solution(iii):

No. of face cards $=8$ ....... (2 red queen, 2 black jack (face cards) already drawn)

Therefore, ${}_8{C}_1$( Selecting $1$ out of $8$ items) times out of ${}^{48}{C}_1$( Selecting $1$ out of $48$ items) a face card is picked.

Let $E$ be the event of getting a face card from the pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^8{C}_1}{{}^{48}{C}_1}$$=\frac{1}{6}$

Solution(iv):

No. of queen cards $=4-2=2$...(2 red queen cards already drawn)

Therefore, ${}^2{C}_1$( Selecting $1$ out of $2$ items) times out of ${}^{48}{C}_1$( Selecting $1$ out of $48$ items) a queen is picked.

Let $E$ be the event of getting a queen from the pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^2{C}_1}{{}^{48}{C}_1}$$=\frac{1}{24}$