Binomial Probability Worksheet with Key | Exercises Probability ... - Free Printable

Let's solve each problem step-by-step using binomial probability concepts.

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Binomial Probability Basics

For a binomial distribution:
- $ n $: number of trials
- $ p $: probability of success on a single trial
- $ q = 1 - p $: probability of failure
- Mean (expected value): $ \mu = np $
- Standard deviation: $ \sigma = \sqrt{np(1-p)} $
- Probability of exactly $ k $ successes:
$$
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
$$

We'll use this formula and properties to answer all parts.

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## Problems 1–4: Mean, Std Dev, and Probabilities

1. $ n = 12, p = 0.2 $

- Mean: $ \mu = np = 12 \times 0.2 = 2.4 $
- Std dev: $ \sigma = \sqrt{12 \times 0.2 \times 0.8} = \sqrt{1.92} \approx 1.3856 $

> No specific probability asked → just mean and std dev.

✔ Answer:
Mean = 2.4, Std dev ≈ 1.386

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2. $ n = 20, p = 0.5 $

- Mean: $ 20 \times 0.5 = 10 $
- Std dev: $ \sqrt{20 \times 0.5 \times 0.5} = \sqrt{5} \approx 2.236 $

✔ Answer:
Mean = 10, Std dev ≈ 2.236

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3. $ n = 11, p = 0.05 $, find $ P(3 \text{ failures}) $

First, note:
- "Failures" means not success.
- $ p = 0.05 $ is probability of success, so $ q = 1 - p = 0.95 $ is failure.
- So we want $ P(\text{3 failures}) = P(\text{8 successes}) $? Wait — no!

Wait:
If there are 11 trials, and we want 3 failures, that means:
- Number of successes = $ 11 - 3 = 8 $

So we need $ P(X = 8) $ where $ X \sim \text{Bin}(11, 0.05) $

But wait: $ p = 0.05 $ is very small, so getting 8 successes is extremely unlikely.

$$
P(X = 8) = \binom{11}{8} (0.05)^8 (0.95)^3
$$

Calculate:

- $ \binom{11}{8} = \binom{11}{3} = 165 $
- $ (0.05)^8 = 3.90625 \times 10^{-11} $
- $ (0.95)^3 \approx 0.857375 $

So:
$$
P = 165 \times 3.90625 \times 10^{-11} \times 0.857375 \approx 5.46 \times 10^{-9}
$$

This is extremely small, essentially zero.

Alternatively, perhaps the question meant:
"Find $ P(3 \text{ failures}) $" meaning 3 failures, i.e., 8 successes? But with $ p = 0.05 $, it's highly improbable.

But let’s double-check interpretation.

Actually, if $ p = 0.05 $ is success, then failure is $ q = 0.95 $. So 3 failures = 3 events with $ q $, which means 8 successes.

So yes, $ P(X = 8) $ where $ X $ is number of successes.

But this is negligible.

Alternatively, maybe the question wants 3 failures as number of failures, so define $ Y $ = number of failures → $ Y \sim \text{Bin}(11, 0.95) $

Then $ P(Y = 3) = \binom{11}{3} (0.95)^3 (0.05)^8 $

Same calculation as above → still tiny.

So either way, same result.

✔ Answer: $ P(3 \text{ failures}) \approx 5.46 \times 10^{-9} $ → practically 0

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4. $ n = 6, p = 0.35 $, find $ P(\text{at least 3 successes}) $

We want $ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6) $

Use binomial formula:

$$
P(X = k) = \binom{6}{k} (0.35)^k (0.65)^{6-k}
$$

Compute each:

#### $ P(X=3) $
- $ \binom{6}{3} = 20 $
- $ (0.35)^3 = 0.042875 $
- $ (0.65)^3 = 0.274625 $
- $ P = 20 \times 0.042875 \times 0.274625 \approx 20 \times 0.01176 \approx 0.2352 $

#### $ P(X=4) $
- $ \binom{6}{4} = 15 $
- $ (0.35)^4 = 0.01500625 $
- $ (0.65)^2 = 0.4225 $
- $ P = 15 \times 0.01500625 \times 0.4225 \approx 15 \times 0.00634 \approx 0.0951 $

#### $ P(X=5) $
- $ \binom{6}{5} = 6 $
- $ (0.35)^5 = 0.0052521875 $
- $ (0.65)^1 = 0.65 $
- $ P = 6 \times 0.0052521875 \times 0.65 \approx 6 \times 0.003414 \approx 0.0205 $

#### $ P(X=6) $
- $ \binom{6}{6} = 1 $
- $ (0.35)^6 = 0.001838265625 $
- $ P \approx 0.00184 $

Now sum:
$$
P(X \geq 3) \approx 0.2352 + 0.0951 + 0.0205 + 0.00184 = 0.3526
$$

✔ Answer: $ P(\text{at least 3 successes}) \approx 0.353 $

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## Problem 5: Multiple Choice Quiz

$ n = 10 $ questions.

Each has 5 options → guess: $ p = 1/5 = 0.2 $

Diana studies → $ p = 0.75 $

a) Colin passes if he guesses

Assume passing = at least 6 correct? Usually 50% or 60%. But since not specified, assume passing = 6 or more correct.

So $ P(X \geq 6) $ for $ X \sim \text{Bin}(10, 0.2) $

We compute:

$$
P(X \geq 6) = \sum_{k=6}^{10} \binom{10}{k} (0.2)^k (0.8)^{10-k}
$$

But $ p = 0.2 $ is low → these probabilities are small.

We can calculate:

Use calculator or approximate:

But let's compute:

- $ P(X=6) = \binom{10}{6}(0.2)^6(0.8)^4 = 210 \times 0.000064 \times 0.4096 \approx 210 \times 0.00002621 \approx 0.0055 $
- $ P(X=7) = \binom{10}{7}(0.2)^7(0.8)^3 = 120 \times 0.0000128 \times 0.512 \approx 120 \times 0.00000655 \approx 0.000786 $
- Higher terms even smaller.

So total $ P(X \geq 6) \approx 0.0055 + 0.0008 = 0.0063 $

✔ a) $ P(\text{Colin passes}) \approx 0.0063 $ → about 0.63%

b) Diana: $ p = 0.75 $, $ P(X \geq 6) $

We compute $ P(X \geq 6) = 1 - P(X \leq 5) $

Use cumulative binomial:

Or calculate directly:

Use binomial formula or known values.

We’ll compute:

$$
P(X=k) = \binom{10}{k} (0.75)^k (0.25)^{10-k}
$$

But easier to use technology or approximation.

Alternatively, use cumulative table or compute:

We can use:

- $ P(X=6) = \binom{10}{6} (0.75)^6 (0.25)^4 = 210 \times 0.1779785 \times 0.00390625 \approx 210 \times 0.000694 \approx 0.1457 $
- $ P(X=7) = \binom{10}{7} (0.75)^7 (0.25)^3 = 120 \times 0.1334838867 \times 0.015625 \approx 120 \times 0.002054 \approx 0.2465 $
- $ P(X=8) = \binom{10}{8} (0.75)^8 (0.25)^2 = 45 \times 0.100112915 \times 0.0625 \approx 45 \times 0.006257 \approx 0.2815 $
- $ P(X=9) = \binom{10}{9} (0.75)^9 (0.25)^1 = 10 \times 0.075084686 \times 0.25 \approx 10 \times 0.01877 \approx 0.1877 $
- $ P(X=10) = (0.75)^{10} \approx 0.0563 $

Sum:
$ 0.1457 + 0.2465 + 0.2815 + 0.1877 + 0.0563 = 0.9177 $

✔ b) $ P(\text{Diana passes}) \approx 0.918 $

c) Expected number of correct guesses for Colin

$ E[X] = np = 10 \times 0.2 = 2 $

✔ c) 2

d) Expected number for Diana

$ E[X] = 10 \times 0.75 = 7.5 $

✔ d) 7.5

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## Problem 6: Manufacturing Defects

$ n = 50 $, $ p = 0.01 $ (since 99% defect-free → 1% defective)

Let $ X $ = number of defective components

$ X \sim \text{Bin}(50, 0.01) $

a) $ P(\text{none defective}) = P(X = 0) $

$$
P(X=0) = (1 - p)^n = (0.99)^{50} \approx e^{-0.5} \approx 0.6065 \quad (\text{approx})
$$

More accurately: $ (0.99)^{50} \approx 0.6050 $

✔ a) ≈ 0.605

b) $ P(\text{at least one defective}) = 1 - P(X=0) = 1 - 0.605 = 0.395 $

✔ b) ≈ 0.395

c) $ P(\text{at least two defective}) = 1 - P(X=0) - P(X=1) $

We already have $ P(X=0) = 0.605 $

Now $ P(X=1) = \binom{50}{1} (0.01)^1 (0.99)^{49} = 50 \times 0.01 \times (0.99)^{49} $

$ (0.99)^{49} \approx 0.6106 $

So:
$$
P(X=1) = 50 \times 0.01 \times 0.6106 = 0.5 \times 0.6106 = 0.3053
$$

Then:
$$
P(X \geq 2) = 1 - 0.605 - 0.3053 = 1 - 0.9103 = 0.0897
$$

✔ c) ≈ 0.0897

d) Expected number of defective parts

$ E[X] = np = 50 \times 0.01 = 0.5 $

✔ d) 0.5

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## Problem 7: Cracked Eggs

$ n = 24 $ (two dozen), $ p = 0.03 $ (3% cracked)

Let $ X $ = number of cracked eggs

$ X \sim \text{Bin}(24, 0.03) $

a) $ P(\text{none cracked}) = P(X=0) = (1 - p)^n = (0.97)^{24} $

Calculate:
$$
(0.97)^{24} \approx e^{-24 \times 0.03} = e^{-0.72} \approx 0.486
$$

More accurately: $ (0.97)^{24} \approx 0.485 $

✔ a) ≈ 0.485

b) $ P(\text{at least one cracked}) = 1 - P(X=0) = 1 - 0.485 = 0.515 $

✔ b) ≈ 0.515

c) $ P(\text{exactly two cracked}) = P(X=2) $

$$
P(X=2) = \binom{24}{2} (0.03)^2 (0.97)^{22}
$$

- $ \binom{24}{2} = 276 $
- $ (0.03)^2 = 0.0009 $
- $ (0.97)^{22} \approx e^{-22 \times 0.03} = e^{-0.66} \approx 0.517 $
- More accurately: $ (0.97)^{22} \approx 0.517 $

So:
$$
P = 276 \times 0.0009 \times 0.517 \approx 276 \times 0.000465 \approx 0.128
$$

✔ c) ≈ 0.128

d) Expected number of cracked eggs

$ E[X] = np = 24 \times 0.03 = 0.72 $

✔ d) 0.72

e) Expected number of uncracked eggs

Total = 24, so expected uncracked = $ 24 - 0.72 = 23.28 $

✔ e) 23.28

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## Problem 8: Tim’s Foul Shots

$ n = 10 $, $ p = 0.7 $

Let $ X $ = number of made shots

$ X \sim \text{Bin}(10, 0.7) $

a) $ P(X = 8) $

$$
P(X=8) = \binom{10}{8} (0.7)^8 (0.3)^2 = 45 \times (0.057648) \times (0.09)
$$

- $ (0.7)^8 \approx 0.057648 $
- $ (0.3)^2 = 0.09 $
- $ 45 \times 0.057648 \times 0.09 \approx 45 \times 0.005188 \approx 0.2335 $

✔ a) ≈ 0.2335

b) $ P(X \geq 8) = P(X=8) + P(X=9) + P(X=10) $

We already have $ P(X=8) \approx 0.2335 $

- $ P(X=9) = \binom{10}{9} (0.7)^9 (0.3)^1 = 10 \times 0.0403536 \times 0.3 \approx 10 \times 0.012106 \approx 0.1211 $
- $ P(X=10) = (0.7)^{10} \approx 0.0282 $

Sum: $ 0.2335 + 0.1211 + 0.0282 = 0.3828 $

✔ b) ≈ 0.383

c) $ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) $

- $ P(X=0) = (0.3)^{10} \approx 5.9049 \times 10^{-6} $
- $ P(X=1) = \binom{10}{1} (0.7)^1 (0.3)^9 = 10 \times 0.7 \times 0.000019683 \approx 10 \times 0.00001378 \approx 0.0001378 $
- $ P(X=2) = \binom{10}{2} (0.7)^2 (0.3)^8 = 45 \times 0.49 \times 0.00006561 \approx 45 \times 0.00003215 \approx 0.001447 $

Sum ≈ $ 0.0000059 + 0.000138 + 0.001447 \approx 0.00159 $

✔ c) ≈ 0.0016

d) $ P(5 \leq X \leq 7) = P(X=5) + P(X=6) + P(X=7) $

Use binomial formula:

- $ P(X=5) = \binom{10}{5} (0.7)^5 (0.3)^5 = 252 \times 0.16807 \times 0.00243 \approx 252 \times 0.000410 \approx 0.103 $
- $ P(X=6) = \binom{10}{6} (0.7)^6 (0.3)^4 = 210 \times 0.117649 \times 0.0081 \approx 210 \times 0.000943 \approx 0.198 $
- $ P(X=7) = \binom{10}{7} (0.7)^7 (0.3)^3 = 120 \times 0.0823543 \times 0.027 \approx 120 \times 0.002223 \approx 0.2668 $

Sum: $ 0.103 + 0.198 + 0.2668 = 0.5678 $

✔ d) ≈ 0.568

e) Expected number of sinks

$ E[X] = np = 10 \times 0.7 = 7 $

✔ e) 7

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## ✔ Final Answers Summary

1.

Mean = 2.4, Std dev ≈ 1.386

2.

Mean = 10, Std dev ≈ 2.236

3.

$ P(3 \text{ failures}) \approx 5.46 \times 10^{-9} \approx 0 $

4.

$ P(\text{at least 3 successes}) \approx 0.353 $

5.

a) $ P(\text{Colin passes}) \approx 0.0063 $
b) $ P(\text{Diana passes}) \approx 0.918 $
c) Expected for Colin: 2
d) Expected for Diana: 7.5

6.

a) $ P(\text{none defective}) \approx 0.605 $
b) $ P(\text{at least one}) \approx 0.395 $
c) $ P(\text{at least two}) \approx 0.0897 $
d) Expected defective: 0.5

7.

a) $ P(\text{none cracked}) \approx 0.485 $
b) $ P(\text{at least one}) \approx 0.515 $
c) $ P(\text{exactly two}) \approx 0.128 $
d) Expected cracked: 0.72
e) Expected uncracked: 23.28

8.

a) $ P(\text{exactly 8}) \approx 0.2335 $
b) $ P(\text{at least 8}) \approx 0.383 $
c) $ P(\text{at most 2}) \approx 0.0016 $
d) $ P(5 \leq X \leq 7) \approx 0.568 $
e) Expected sinks: 7

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