Solved WORKSHEET 11: Two Rules of Probability Name: $ 9. | Chegg.com - Free Printable

Explanation:
We are working with a standard 52-card deck. Let’s recall some key facts:

- 52 total cards
- 4 suits: Hearts (♥), Diamonds (♦), Clubs (♣), Spades (♠)
- Each suit has 13 cards: Ace, 2–10, Jack, Queen, King
- Red cards: Hearts and Diamonds → 26 red cards
- Black cards: Clubs and Spades → 26 black cards
- Face cards: Jack, Queen, King → 3 per suit × 4 suits = 12 face cards
- Aces: 4 total (one per suit)

Now let’s solve each part carefully.

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i. Two cards selected *with replacement* — probability both are Aces

With replacement means after drawing the first card, we put it back, so the deck is full again for the second draw.

- P(first card is Ace) = 4/52 = 1/13
- P(second card is Ace) = also 4/52 = 1/13
- Since independent: P(both Aces) = (1/13) × (1/13) = 1/169

✔ Check: 4/52 = 1/13, yes. Square = 1/169.

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j. Three cards with replacement — all three are Hearts

- Hearts = 13 cards
- P(one card is Heart) = 13/52 = 1/4
- With replacement: independent draws
- P(all three Hearts) = (1/4)³ = 1/64

✔ Correct.

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k. Three cards with replacement — all are *not* Face cards

- Total face cards = 12 (J, Q, K × 4 suits)
- So non-face cards = 52 − 12 = 40
- P(one card is not a face card) = 40/52 = 10/13
- With replacement: (10/13)³ = 1000 / 2197

Let’s compute: 10³ = 1000, 13³ = 2197 → yes.

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l. Two cards *without* replacement — both are Red cards

- Red cards = 26
- First draw: P(red) = 26/52 = 1/2
- After removing one red card: 25 red left, 51 total cards
- Second draw: P(red | first was red) = 25/51
- Joint probability: (26/52) × (25/51) = (1/2) × (25/51) = 25/102

Simplify? 25 and 102 share no common factor (25 = 5², 102 = 2×3×17), so 25/102 is final.

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m. Two cards without replacement — both are Aces

- Aces = 4
- First draw: 4/52 = 1/13
- After drawing one Ace: 3 Aces left, 51 cards total
- Second draw: 3/51 = 1/17
- Multiply: (4/52) × (3/51) = (1/13) × (1/17) = 1/221

Check: 4×3 = 12, 52×51 = 2652 → 12/2652 = divide numerator & denominator by 12 → 1/221. ✔

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n. Three cards without replacement — all three are Hearts

- Hearts = 13
- First: 13/52 = 1/4
- Second: 12/51
- Third: 11/50
- Multiply: (13/52) × (12/51) × (11/50)

Simplify step by step:

13/52 = 1/4
So: (1/4) × (12/51) × (11/50)
= (12 × 11) / (4 × 51 × 50)
= 132 / (4 × 2550) = 132 / 10200
Wait—better to reduce earlier:

13/52 × 12/51 × 11/50
Cancel 13 and 52: 13/52 = 1/4
Now: (1/4) × (12/51) × (11/50)
12 and 4: 12/4 = 3 → becomes (3/1) × (1/51) × (11/50) = (3 × 11) / (51 × 50) = 33 / 2550
Now simplify 33/2550: divide numerator & denominator by 3 → 11 / 850

Check: 13×12×11 = 1716
52×51×50 = 132600
1716 / 132600 → divide numerator & denominator by 156? Let’s do GCD:

1716 ÷ 12 = 143
132600 ÷ 12 = 11050
143 / 11050 → 143 = 11×13, 11050 = 11×1004.545… no.

Better: use calculator-style simplification:

13/52 = 1/4
12/51 = 4/17 (divide numerator & denominator by 3)
11/50 stays.

So: (1/4) × (4/17) × (11/50) = (1 × 4 × 11) / (4 × 17 × 50)
Cancel 4 top and bottom → 11 / (17 × 50) = 11 / 850

Yes! So answer is 11/850

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o. Three cards without replacement — all are *not* Face cards

Non-face cards = 52 − 12 = 40
We draw 3 without replacement.

P = (40/52) × (39/51) × (38/50)

Simplify step by step:

40/52 = 10/13
39/51 = 13/17 (divide by 3)
38/50 = 19/25

So multiply: (10/13) × (13/17) × (19/25)
Cancel 13 top and bottom: → (10/1) × (1/17) × (19/25) = (10 × 19) / (17 × 25) = 190 / 425

Simplify: divide numerator and denominator by 5 → 38 / 85

Check: 190 ÷ 5 = 38, 425 ÷ 5 = 85. Yes.

So answer: 38/85

Double-check original product:
40×39×38 = 40×1482 = 59280
52×51×50 = 132600
59280 / 132600 → divide both by 1560? Let’s divide numerator and denominator by 1560:

132600 ÷ 1560 = 85
59280 ÷ 1560 = 38
Yes! So 38/85 is correct.

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p. Two cards without replacement — probability second card is red *given* first card is black

This is a conditional probability: P(second is red | first is black)

Given first card is black → we removed one black card.

Original: 26 black, 26 red
After removing 1 black: 25 black, 26 red, total 51 cards left.

So P(second is red | first black) = 26 / 51

That’s it — no need to multiply anything else.

✔ Final check: yes, conditional probability only depends on remaining deck.

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Now compile all answers:

i. 1/169
j. 1/64
k. 1000/2197
l. 25/102
m. 1/221
n. 11/850
o. 38/85
p. 26/51

All verified.

Final Answer:
i. $\frac{1}{169}$
j. $\frac{1}{64}$
k. $\frac{1000}{2197}$
l. $\frac{25}{102}$
m. $\frac{1}{221}$
n. $\frac{11}{850}$
o. $\frac{38}{85}$
p. $\frac{26}{51}$