CRE 3 - Pack of Cards

When a card is drawn total possible outcomes = 52.

For a card to be either black or queen, total desired outcomes = 26 + 2 = 28.

(26 black cards and 2 red queens)

∴ Required probability = 28/52 = 7/13.

Hence, option (c).

When a card is drawn total possible outcomes = ⁵²C₂

No of ways of picking 1 red and 1 black card = ²⁶C₁ × ²⁶C₁

∴ Required probability = (26C1 × 26C1)/ 52C2 = $\frac{\frac{26\times 26}{52\times 51}}{2}$ = 26/51.

Hence, option (a).

P (picking at least 1 honor card) = 1 – P (not picking any honor card)

When a card is drawn total possible outcomes = ⁵²C₂

There are total 20 honor cards in a pack of cards. 

∴ No of ways of not picking any honor card = ³⁶C₂

∴ P (not picking any honor card) = ³⁶C₂ / ⁵²C₂ = $\frac{36\times 35}{52\times 51}$ = $\frac{105}{221}$.

P (picking at least 1 honor card) = 1 – $\frac{105}{221}$ = $\frac{116}{221}$

Hence, option (b).

When two cards are drawn total possible outcomes = ⁵²C₂

Total cards which are either black or queen = 28.

Number of ways of drawing neither black nor queen = ²⁴C₂

∴ Number of ways of drawing either black or queen = ⁵²C₂ - ²⁴C₂

∴ P (picking either a black card or a queen) = (⁵²C₂ - ²⁴C₂)/ ⁵²C₂ = 1 - $\frac{24\times 23}{52\times 51}$ = 1 - $\frac{46}{221}$ = $\frac{175}{221}$

Hence, option (c).

Either the first card can be red and second black or vice versa.

P (1^st red) = 26/52 = 1/2.

P (2^nd black) = 26/52 = 1/2

∴ P (1^st red and 2nd black) = 1/2 × 1/2 = 1/4

∴ P (1^st black and 2nd red) = 26/52 × 26/52 = 1/4

P (a red and a black card) = P (1^st red and 2^nd black) + P (1^st black and 2^nd red) = 1/2.

Hence, option (b).

P (picking at least 1 honor card) = 1 – P (not picking any honor card)

There are total 20 honor cards in a pack of cards i.e., 36 non-honor cards.

P (not picking honor card in 1^st draw) = 36/52 = 9/13

Now, P (not picking honor card in 2^nd draw) = 36/52 = 9/13

∴ P (not picking any honor card) = 9/13 × 9/13 = 81/169

∴ P (picking at least 1 honor card) = 1 – 81/169 = 88/169

Hence, option (c).

P (picking either a black or a queen) = 1 – P (picking neither a black nor a queen)

Total cards which are either black or queen = 28.

∴ P (picking neither a black nor a queen) = 24/52 × 24/52 = 36/169

∴ P (picking either a black or a queen) = 1 – 36/169 = 133/169

Hence, option (d).

Either the first card can be red and second black or vice versa.

P (1^st red) = 26/52 = 1/2.

P (2^nd black) = 26/51

∴ P (1^st red and 2^nd black) = 1/2 × 26/51 = 13/51

∴ P (1^st black and 2^nd red) = 26/52 × 26/51 = 13/51

P (a red and a black card) = P (1^st red and 2^nd black) + P (1^st black and 2^nd red) = 26/51.

Hence, option (d).

P (picking at least 1 honor card) = 1 – P (not picking any honor card)

There are total 20 honor cards in a pack of cards i.e., 36 non-honor cards.

P (not picking honor card in 1^st draw) = 36/52 = 9/13

Now, P (not picking honor card in 2^nd draw) = 35/51

∴ P (not picking any honor card) = 9/13 × 35/51 = 105/221

∴ P (picking at least 1 honor card) = 1 – 105/221 = 116/221

Hence, option (c).

P (picking either a black or a queen) = 1 – P (picking neither a black nor a queen)

Total cards which are either black or queen = 28.

∴ P (picking neither a black nor a queen) = 24/52 × 23/51 = 46/221

∴ P (picking either a black or a queen) = 1 – 46/221 = 175/221

Hence, option (a).