Question

All the black face cards are removed from a pack of $52$ cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a
(i) face card (ii) red card (iii) black card (iv) king

Answer 1

As there are $6$ black face cards in a pack $52$ cards, and those cards are removed so,
No. of cards in pack $=(52-6)=46$

Solution(i):

No. of face cards $=12-6=6$....(6 black face cards are removed)

Therefore, ${}^6{C}_1$( Selecting $1$ out of $6$ items) times out of ${}^{46}{C}_1$( Selecting $1$ out of $46$ 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)}}$$=\frac{{}^6{C}_1}{{}^{46}{C}_1}$$=\frac{3}{23}$

Solution(ii):

No. of red cards $=26$

Therefore, ${}^{26}{C}_1$( Selecting $1$ out of $26$ items) times out of ${}^{46}{C}_1$( Selecting $1$ out of $46$ 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)}}$$=\frac{{}^{26}{C}_1}{{}^{46}{C}_1}$$=\frac{13}{23}$

Solution(iii):

No. of black cards $=26-6=20$......(As 6 black cards aare removed already)

Therefore, ${}^{20}{C}_1$( Selecting $1$ out of $20$ items) times out of ${}^{46}{C}_1$( Selecting $1$ out of $46$ items) a black card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{20}{C}_1}{{}^{46}{C}_1}$$=\frac{10}{23}$

Solution(iv):

No. of kings $=4-2=2$ ......(As 2 black king card (face card) are removed already)

Therefore, ${}^2{C}_1$( Selecting $1$ out of $2$ items) times out of ${}^{46}{C}_1$( Selecting $1$ out of $46$ items) a king is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^2{C}_1}{{}^{46}{C}_1}$$=\frac{1}{23}$