Probability Practice 2.pdf - Free Printable

Let’s solve each part step by step.

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Problem 1: Socks in a drawer

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

We are drawing without replacement, so the total number of socks decreases with each draw.

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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) = 1/7 ≈ 0.1429

✔ Check: 6×5=30, 15×14=210 → simplify 30/210 = 1/7 ✔️

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b) Three socks drawn — probability all three are white

White socks = 8

First draw: 8/15
Second: 7/14
Third: 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

✔ Check: 48 ÷ 6 = 8, 390 ÷ 6 = 65 ✔️

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c) Two socks drawn — probability both are white

Same as above but only two draws:

(8/15) × (7/14) = (8/15) × (1/2) = 8/30 = 4/15

✔ Check: 8×7=56, 15×14=210 → 56/210 = 8/30 = 4/15 ✔️

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d) Four socks drawn — probability that a pair of white socks will be drawn

This is tricky. “A pair of white socks” means at least two white socks among the four drawn.

It’s easier to calculate the opposite: probability of getting fewer than 2 white socks (i.e., 0 or 1 white), then subtract from 1.

Total ways to choose 4 socks from 15: C(15,4) — we’ll use combinations.

But since we’re dealing with probabilities without replacement, let’s think in terms of hypergeometric distribution.

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

We want P(at least 2 white) = 1 - [P(0 white) + P(1 white)]

Calculate P(0 white): all 4 socks from non-white (7 socks)

C(7,4) / C(15,4)

C(7,4) = 35
C(15,4) = 1365
→ P(0 white) = 35/1365

P(1 white): choose 1 white from 8, and 3 non-white from 7

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

So P(fewer than 2 white) = (35 + 280)/1365 = 315/1365

Simplify: divide numerator and denominator by 105 → 3/13

Wait: 315 ÷ 105 = 3, 1365 ÷ 105 = 13? Let me check: 105×13=1365 ✔️, 105×3=315 ✔️

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

✔ But wait — let me double-check the combination values.

C(15,4) = 15×14×13×12 / (4×3×2×1) = (32760)/24 = 1365 ✔️
C(7,4) = 35 ✔️
C(7,3) = 35 ✔️
C(8,1)=8 ✔️
So 8×35=280 ✔️
35+280=315 ✔️
315/1365 = 315÷105=3, 1365÷105=13 → 3/13 ✔️
1 - 3/13 = 10/13 ✔️

So answer is 10/13.

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

There is only 1 purple sock in the drawer.

You cannot draw a pair (two) of purple socks if there’s only one!

So probability = 0

✔ Simple logic: impossible event → probability 0 ✔️

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Problem 2: Marbles in a bag

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

All equally likely to be drawn.

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a) Probability of drawing a blue marble (one draw)

P(blue) = 8/23

That’s it. No simplification needed.

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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 step by step:

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

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: 36÷3=12, 759÷3=253 ✔️

Is 12 and 253 divisible by anything? 253 ÷ 11 = 23, 12 not divisible by 11 or 23 → so 12/253 is simplified.

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

Again, use complement: 1 - P(both yellow)

Yellow = 6

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

Simplify: divide by 2 → 15/253

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

Check if can simplify: 238 and 253.

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

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Now, compiling final answers clearly.

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