Question

From a pack of $52$ playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is
(i) a black queen (ii) a red card (iii) a black jack (iv) a picture card (Jacks, queens and kings are picture cards)

Answer 1

Total no. of cards in a pack $=52$

After removing red-colored - jack, queen, king, and aces,

No. of cards $=52-8=44$ and

No. of red cards $=26-8=18$.

Solution(i):

No. of black queens $=2$

Therefore, ${}^2{C}_1$( Selecting $1$ out of $2$ items) times out of ${}^{44}{C}_1$( Selecting $1$ out of $44$ items) a black queen is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^2{C}_1}{{}^{44}{C}_1}$$=\frac{2}{44}$$=\frac{1}{22}$

Solution(ii):

No. of red cards $=18$

Therefore, ${}^{18}{C}_1$( Selecting $1$ out of $18$ items) times out of ${}^{44}{C}_1$( Selecting $1$ out of $44$ items) a red card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{18}{C}_1}{{}^{44}{C}_1}$$=\frac{18}{44}$$=\frac{9}{22}$

Solution(iii):

No. of black jack cards $=2$

Therefore, ${}^2{C}_1$( Selecting $1$ out of $2$ items) times out of ${}^{44}{C}_1$( Selecting $1$ out of $44$ items) a black jack card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^2{C}_1}{{}^{44}{C}_1}$$=\frac{2}{44}$$=\frac{1}{22}$

Solution(iv):

No. of picture cards $=12-6=6$ ..... (As red cards are removed and there are 6 red picture cards )

Therefore, ${}^6{C}_1$( Selecting $1$ out of $6$ items) times out of ${}^{44}{C}_1$( Selecting $1$ out of $44$ items) a picture card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^6{C}_1}{{}^{44}{C}_1}$$=\frac{6}{44}$$=\frac{3}{22}$