Probability Worksheet 2 featuring problems on calculating probabilities of drawing specific combinations of socks and marbles without replacement.

Let’s solve each part step by step. We’ll use basic probability rules:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)

We are drawing without replacement, so the total number of items decreases after each draw, and we must adjust probabilities accordingly.

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Problem 1: Socks in a drawer

Total socks = 15
Blue = 6, White = 8, Purple = 1

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#### a) Two socks drawn — probability both are blue?

First draw: P(blue) = 6/15
Second draw (one blue already taken): P(blue) = 5/14

Multiply them:
(6/15) × (5/14) = (30)/(210) = simplify → divide numerator and denominator by 30 → 1/7

✔ Answer for 1a: 1/7

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#### b) Three socks drawn — probability all three are white?

White socks = 8

First draw: P(white) = 8/15
Second draw: P(white) = 7/14
Third draw: P(white) = 6/13

Multiply:
(8/15) × (7/14) × (6/13)

Simplify step by step:

7/14 = 1/2 → so now: (8/15) × (1/2) × (6/13) = (8 × 1 × 6) / (15 × 2 × 13) = 48 / 390

Simplify 48/390: divide numerator and denominator by 6 → 8/65

✔ Answer for 1b: 8/65

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#### c) Two socks drawn — probability both are white?

Same as above but only two draws.

P(first white) = 8/15
P(second white) = 7/14 = 1/2

Multiply: (8/15) × (1/2) = 8/30 = 4/15

✔ Answer for 1c: 4/15

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#### d) Four socks drawn — probability that *at least one pair* of white socks is drawn?

Wait — this wording is tricky. “A pair of white socks” usually means two white socks. But if you draw four socks, getting “a pair” could mean exactly two, or at least two? In context, since it says “a pair”, and earlier parts asked for “all blue” or “all white”, here it likely means at least two white socks (i.e., forming at least one pair).

But let’s read carefully: “what is the probability that a pair of white socks will be drawn?” — meaning, among the 4 drawn, there are at least 2 white ones.

So we want: P(at least 2 white socks in 4 drawn)

It’s easier to calculate the complement:
P(no white or only 1 white) = P(0 white) + P(1 white)
Then subtract from 1.

Total ways to choose 4 socks from 15: C(15,4) — but since we’re doing probability with sequential draws, we can also do it step-by-step using multiplication, but combinations might be cleaner.

Actually, since order doesn’t matter for “having a pair”, let’s use combinations.

Total ways to choose 4 socks from 15: C(15,4) = 1365

Favorable: at least 2 white socks.

White socks = 8, non-white = 7 (6 blue + 1 purple)

Cases:
- Exactly 2 white, 2 non-white: C(8,2) × C(7,2)
- Exactly 3 white, 1 non-white: C(8,3) × C(7,1)
- Exactly 4 white, 0 non-white: C(8,4) × C(7,0)

Calculate:

C(8,2) = 28, C(7,2) = 21 → 28×21 = 588
C(8,3) = 56, C(7,1) = 7 → 56×7 = 392
C(8,4) = 70, C(7,0) = 1 → 70×1 = 70

Total favorable = 588 + 392 + 70 = 1050

Total possible = C(15,4) = 1365

So probability = 1050 / 1365

Simplify: divide numerator and denominator by 105 → 1050÷105=10, 1365÷105=13 → wait, 105×13=1365? 105×13=1365 yes. 105×10=1050. So 10/13? Wait no:

Wait 1050 ÷ 105 = 10? 105×10=1050 yes. 1365÷105=13? 105×13=1365 yes. So 10/13? That seems too high.

Wait let me check C(15,4):

C(15,4) = (15×14×13×12)/(4×3×2×1) = (32760)/24 = 1365 ✔️

C(8,2)=28, C(7,2)=21 → 28×21=588 ✔️
C(8,3)=56, C(7,1)=7 → 392 ✔️
C(8,4)=70 ✔️
Sum: 588+392=980; 980+70=1050 ✔️

1050 / 1365

Divide numerator and denominator by 15: 1050÷15=70, 1365÷15=91 → 70/91

Divide by 7: 10/13 → same as before.

But 10/13 ≈ 0.769 — is that reasonable? Let’s think: there are 8 white out of 15, so more than half. Drawing 4 socks, chance of getting at least 2 white should be high. Yes, plausible.

Alternatively, compute complement:

P(0 white) = C(7,4)/C(15,4) = 35 / 1365
P(1 white) = C(8,1)×C(7,3)/C(15,4) = 8 × 35 / 1365 = 280 / 1365

C(7,3)=35, yes.

So P(0 or 1 white) = (35 + 280)/1365 = 315/1365

Simplify: divide by 105 → 3/13

Then P(at least 2 white) = 1 - 3/13 = 10/13

Yes! Confirmed.

✔ Answer for 1d: 10/13

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#### e) Eight socks drawn — probability that a pair of purple socks will be drawn?

Purple socks = only 1!

You cannot draw a pair (which requires 2) of purple socks if there’s only 1 purple sock.

So impossible.

✔ Answer for 1e: 0

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Problem 2: Marbles in a bag

Blue = 8, Orange = 9, Yellow = 6 → Total = 23 marbles

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#### a) Probability of drawing a blue marble?

Just one draw.

P(blue) = 8/23

✔ Answer for 2a: 8/23

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#### b) Three marbles drawn — probability all are orange?

Orange = 9

First draw: 9/23
Second: 8/22
Third: 7/21

Multiply: (9/23) × (8/22) × (7/21)

Simplify:

8/22 = 4/11
7/21 = 1/3

So: (9/23) × (4/11) × (1/3) = (9 × 4 × 1) / (23 × 11 × 3) = 36 / 759

Simplify: divide numerator and denominator by 3 → 12 / 253

Check if 12 and 253 have common factors: 253 ÷ 11 = 23, so 253=11×23. 12 has no 11 or 23 → simplified.

✔ Answer for 2b: 12/253

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#### c) Two marbles drawn — probability of NOT drawing 2 yellow marbles?

This means: 1 minus the probability of drawing 2 yellow marbles.

Yellow = 6

P(both yellow) = (6/23) × (5/22) = 30 / 506 = 15 / 253

So P(not both yellow) = 1 - 15/253 = (253 - 15)/253 = 238/253

Can we simplify? 238 and 253.

253 = 11×23
238 ÷ 2 = 119, 119=7×17 → no common factors with 253.

So 238/253 is simplified.

✔ Answer for 2c: 238/253

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Final Answers:

1a) 1/7
1b) 8/65
1c) 4/15
1d) 10/13
1e) 0
2a) 8/23
2b) 12/253
2c) 238/253

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Final Answer:
1a) \frac{1}{7}
1b) \frac{8}{65}
1c) \frac{4}{15}
1d) \frac{10}{13}
1e) 0
2a) \frac{8}{23}
2b) \frac{12}{253}
2c) \frac{238}{253}