Question

A card is drawn at random from a pack of $52$ cards. Find the probability that the card is drawn is
(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen

Answer 1

No. of cards in a pack $=52$

Solution(i):

No. of black kings $=2$

Therefore, ${}^2{C}_1$( Selecting $1$ out of $2$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a black king is picked.

Let $E$ be the event of getting a black kingfrom pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^2{C}_1}{{}^{52}{C}_1}$$=\frac{2}{52}$$=\frac{1}{26}$

Solution(ii):

No. of black cards or kings $=28$..... (26Black(including 2 Black kings) + 2 Red Kings)

Therefore, ${}^{28}{C}_1$( Selecting $1$ out of $28$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a either a black card or a king is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{28}{C}_1}{{}^{52}{C}_1}$$=\frac{28}{52}$$=\frac{7}{13}$

Solution(iii):

No. of jack, queen or king $=12$ (4- Jack, 4- Queen, 4-King)

Therefore, ${}^{12}{C}_1$( Selecting $1$ out of $12$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a jack, queen or a king is picked.

Let $E$ be the event of getting ajack, queen or a king from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{12}{C}_1}{{}^{52}{C}_1}$$=\frac{12}{52}$$=\frac{3}{13}$

Solution(iv):

No. of spade or ace $=16$ ........(13-Spade(including 1-Ace) + 3-Aces)

Therefore, ${}^{16}{C}_1$( Selecting $1$ out of $16$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a spade or an ace is picked.

Let $E$ be the event of getting a spade or an ace from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{16}{C}_1}{{}^{52}{C}_1}$$=\frac{16}{52}$$=\frac{4}{13}$

Solution(v):

No. of neither ace nor king$=52-8=44$ ...... (As there are 4-Kings and 4-Aces)

Therefore, ${}^{44}{C}_1$( Selecting $1$ out of $44$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a neither an ace nor a king is picked.

Let $E$ be the event of gettingneither an ace nor a king from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{44}{C}_1}{{}^{52}{C}_1}$$=\frac{44}{52}$$=\frac{11}{13}$

Solution(vi):

No. of neither red nor queen $=52-28=24$ ...(26-red+ 2-Black Queen)

Therefore, ${}^{24}{C}_1$( Selecting $1$ out of $24$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) neither red nor a queen is picked.

Let $E$ be the event of gettingneither red nor queen from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{24}{C}_1}{{}^{52}{C}_1}$$=\frac{24}{52}$$=\frac{6}{13}$

Solution(vii):

No. of non-ace cards $=52-4=$ $48$ ......(4 Aces)

Therefore, ${}^{48}{C}_1$( Selecting $1$ out of $48$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) non-ace card is picked.

Let $E$ be the event of getting a non-ace card from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{48}{C}_1}{{}^{52}{C}_1}$$=\frac{48}{52}$$=\frac{12}{13}$

Solution(viii):

No. of cards with no. $10$ $=4$

Therefore, ${}^4{C}_1$( Selecting $1$ out of $4$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) card with no. $10$ is picked.

Let $E$ be the event of getting a card with no. $10$ from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^4{C}_1}{{}^{52}{C}_1}$$=\frac{4}{52}$$=\frac{1}{13}$

Solution(ix):

No. of spade cards $=13$

Therefore, ${}^{13}{C}_1$( Selecting $1$ out of $13$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a spade card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{13}{C}_1}{{}^{52}{C}_1}$$=\frac{13}{52}$$=\frac{1}{4}$

Solution(x):

No. of black cards $=26$

Therefore, ${}^{26}{C}_1$( Selecting $1$ out of $26$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a black card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{26}{C}_1}{{}^{52}{C}_1}$$=\frac{26}{52}$$=\frac{1}{2}$

Solution(xi):

No. of seven of clubs $=1$

Therefore, ${}^1{C}_1$( Selecting $1$ out of $1$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a seven of clubs is picked

Let $E$ be the event of getting a seven of clubs from the pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^1{C}_1}{{}^{52}{C}_1}$$=\frac{1}{52}$$=\frac{1}{52}$

Solution(xii):

No. of jacks $=4$

Therefore, ${}^4{C}_1$( Selecting $1$ out of $4$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a jack is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^4{C}_1}{{}^{52}{C}_1}$$=\frac{4}{52}$$=\frac{1}{13}$

Solution(xiii):

No. of ace of spades $=1$

Therefore, ${}^1{C}_1$( Selecting $1$ out of $1$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) an ace of spade is picked

Let $E$ be the event of getting an ace of spade from the pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^1{C}_1}{{}^{52}{C}_1}$$=\frac{1}{52}$$=\frac{1}{52}$

Solution(xiv):

No. of queens $=4$

Therefore, ${}^4{C}_1$( Selecting $1$ out of $4$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a queen is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^4{C}_1}{{}^{52}{C}_1}$$=\frac{4}{52}$$=\frac{1}{13}$

Solution(xv):

No. of heart cards $=13$

Therefore, ${}^{13}{C}_1$( Selecting $1$ out of $13$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a heart card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{13}{C}_1}{{}^{52}{C}_1}$$=\frac{13}{52}$$=\frac{1}{4}$

Solution(xvi):

No. of red cards $=26$

Therefore, ${}^{26}{C}_1$( Selecting $1$ out of $26$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a red card is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{26}{C}_1}{{}^{52}{C}_1}$$=\frac{26}{52}$$=\frac{1}{2}$

Solution(xvii):

No. of neither king nor queen cards $=52-8=44$ ...... (As there are 4-Kings and 4-Queen)

Therefore, ${}^{44}{C}_1$( Selecting $1$ out of $44$ items) times out of ${}^{52}{C}_1$( Selecting $1$ out of $52$ items) a neither king nor queen card is picked.

Let $E$ be the event of gettingneither king nor queen card from pack

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{44}{C}_1}{{}^{52}{C}_1}$$=\frac{44}{52}$$=\frac{11}{13}$