# CRE 3 - Pack of Cards | Modern Math - Probability

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**CRE 3 - Pack of Cards | Modern Math - Probability**

A card is drawn at random from a well-shuffled pack of 52 cards. What is the probability that it is either a black card or a queen?

- (a)
27/52

- (b)
1/2

- (c)
7/13

- (d)
None of these

Answer: Option C

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**Explanation** :

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).

Workspace:

**Answer the next 3 questions based on the information given below:**

Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that

**CRE 3 - Pack of Cards | Modern Math - Probability**

One red and one black card?

- (a)
26/51

- (b)
116/221

- (c)
175/221

- (d)
127/221

Answer: Option A

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**Explanation** :

When a card is drawn total possible outcomes = ^{52}C_{2}

No of ways of picking 1 red and 1 black card = ^{26}C_{1} × ^{26}C_{1}

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

Hence, option (a).

Workspace:

**CRE 3 - Pack of Cards | Modern Math - Probability**

At least one honor card?

- (a)
26/51

- (b)
116/221

- (c)
175/221

- (d)
127/221

Answer: Option B

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**Explanation** :

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

When a card is drawn total possible outcomes = ^{52}C_{2}

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

∴ No of ways of not picking any honor card = ^{36}C_{2}

∴ P (not picking any honor card) = ^{36}C_{2 }/ ^{52}C_{2} = $\frac{36\times 35}{52\times 51}$ = $\frac{105}{221}$.

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

Hence, option (b).

Workspace:

**CRE 3 - Pack of Cards | Modern Math - Probability**

Either a black card or a queen is drawn?

- (a)
26/51

- (b)
116/221

- (c)
175/221

- (d)
127/221

Answer: Option C

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**Explanation** :

When two cards are drawn total possible outcomes = ^{52}C_{2}

Total cards which are either black or queen = 28.

Number of ways of drawing neither black nor queen = ^{24}C_{2}

∴ Number of ways of drawing either black or queen = ^{52}C_{2} - ^{24}C_{2}

∴ P (picking either a black card or a queen) = (^{52}C_{2} - ^{24}C_{2})/ ^{52}C_{2} = 1 - $\frac{24\times 23}{52\times 51}$ = 1 - $\frac{46}{221}$ = $\frac{175}{221}$

Hence, option (c).

Workspace:

**Answer the next 3 questions based on the information given below:**

Two cards are drawn at random from a well-shuffled pack of 52 cards one after another with replacement. What is the probability that:

**CRE 3 - Pack of Cards | Modern Math - Probability**

One red and one black card?

- (a)
3/11

- (b)
1/2

- (c)
88/169

- (d)
133/169

Answer: Option B

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**Explanation** :

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).

Workspace:

**CRE 3 - Pack of Cards | Modern Math - Probability**

At least one honor card?

- (a)
3/11

- (b)
1/2

- (c)
88/169

- (d)
133/169

Answer: Option C

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**Explanation** :

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).

Workspace:

**CRE 3 - Pack of Cards | Modern Math - Probability**

Either a black card or a queen is drawn?

- (a)
3/11

- (b)
1/2

- (c)
88/169

- (d)
133/169

Answer: Option D

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**Explanation** :

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).

Workspace:

**Answer the next 3 questions based on the information given below:**

Two cards are drawn at random from a well-shuffled pack of 52 cards one after another without replacement. What is the probability that:

**CRE 3 - Pack of Cards | Modern Math - Probability**

One red and one black card?

- (a)
175/221

- (b)
1/2

- (c)
116/221

- (d)
26/51

Answer: Option D

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**Explanation** :

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).

Workspace:

**CRE 3 - Pack of Cards | Modern Math - Probability**

At least one honor card?

- (a)
175/221

- (b)
1/2

- (c)
116/221

- (d)
26/51

Answer: Option C

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**Explanation** :

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).

Workspace:

**CRE 3 - Pack of Cards | Modern Math - Probability**

Either a black card or a queen is drawn?

- (a)
175/221

- (b)
1/2

- (c)
116/221

- (d)
26/51

Answer: Option A

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**Explanation** :

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).

Workspace:

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