A binomial experiment has the following properties:
The probability of getting exactly \( k \) successes in \( n \) trials is:
\[ P(k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k} \]where the binomial coefficient is:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]and \( n! = 1 \times 2 \times 3 \times \cdots \times n \).
A fair coin is tossed 3 times. Find the probability of getting exactly 2 heads.
Sample space for 3 tosses:
Event \( E \) = {HHT, HTH, THH} has 3 favorable outcomes.
For each outcome with 2 heads and 1 tail:
\[ P(\text{HHT}) = p \cdot p \cdot (1-p) = p^2(1-p) \]Since \( p = 0.5 \) for a fair coin:
\[ P(E) = 3 \times (0.5)^2(0.5) = 3 \times 0.125 = 0.375 \]The coefficient 3 comes from combinations: \(\binom{3}{2} = 3\).
For a binomial distribution:
A fair coin is tossed 5 times. What's the probability of exactly 3 heads?
Here \( n = 5 \), \( k = 3 \), \( p = 0.5 \):
\[ \begin{aligned} P(3 \text{ heads}) &= \binom{5}{3} (0.5)^3 (0.5)^2 \\ &= 10 \times 0.125 \times 0.25 \\ &= 0.3125 \end{aligned} \]A fair die is rolled 7 times. Find the probability of exactly 5 sixes.
Here \( n = 7 \), \( k = 5 \), \( p = \frac{1}{6} \):
\[ P(5 \text{ sixes}) = \binom{7}{5} \left(\frac{1}{6}\right)^5 \left(\frac{5}{6}\right)^2 = 21 \times 0.0001286 \times 0.6944 \approx 0.00187 \]A factory produces tools with 98% reliability. In samples of 1000 tools:
Here \( n = 1000 \), \( p = 0.98 \):
A fair coin is tossed 7 times. Find the probability of at least 5 heads.
"At least 5" means 5, 6, or 7 heads:
\[ \begin{aligned} P(\geq 5) &= P(5) + P(6) + P(7) \\ &= \binom{7}{5}(0.5)^5(0.5)^2 + \binom{7}{6}(0.5)^6(0.5)^1 + \binom{7}{7}(0.5)^7(0.5)^0 \\ &= 0.16406 + 0.05469 + 0.00781 \\ &= 0.22656 \end{aligned} \]A test has 20 questions with 4 choices each. What's the probability of passing (≥10 correct) by guessing?
Here \( n = 20 \), \( p = 0.25 \):
\[ \begin{aligned} P(\geq 10) &= \sum_{k=10}^{20} \binom{20}{k} (0.25)^k (0.75)^{20-k} \\ &\approx 0.00992 + 0.00301 + 0.00075 + \cdots \\ &\approx 0.01386 \quad (\text{about 1.4%}) \end{aligned} \]Conclusion: Random guessing is ineffective for passing.
A box has 3 red, 4 white, and 3 black balls. Six draws with replacement. Find the probability of at least 2 red balls.
Probability of red in one draw: \( p = \frac{3}{10} = 0.3 \)
Use complement rule:
\[ \begin{aligned} P(\geq 2) &= 1 - [P(0) + P(1)] \\ &= 1 - \left[\binom{6}{0}(0.3)^0(0.7)^6 + \binom{6}{1}(0.3)^1(0.7)^5\right] \\ &= 1 - [0.11765 + 0.30253] \\ &= 0.57982 \end{aligned} \]80% of people in a city have insurance with "MyInsurance".
A die is rolled 5 times.
A card is drawn from a deck (with replacement) 10 times.
A multiple-choice test has 20 questions with 5 choices each. What's the probability of passing (≥15 correct) by guessing?
In the 25-34 age group, 61.8% in Canada and 50.8% in the UK have tertiary education. If 200,000 people are randomly selected from each country, how many are expected to have tertiary education?
For a die, probability of even number: \( p = \frac{3}{6} = 0.5 \), \( n = 5 \)
Probability of red card: \( p = \frac{26}{52} = 0.5 \), \( n = 10 \)
Using complement rule:
\[ \begin{aligned} P(\geq 3) &= 1 - [P(0) + P(1) + P(2)] \\ &= 1 - \left[\binom{10}{0}(0.5)^{10} + \binom{10}{1}(0.5)^{10} + \binom{10}{2}(0.5)^{10}\right] \\ &= 1 - [0.00098 + 0.00977 + 0.04395] \\ &= 1 - 0.0547 = 0.9453 \end{aligned} \]Here \( n = 20 \), \( p = 0.2 \):
\[ P(\geq 15) = \sum_{k=15}^{20} \binom{20}{k} (0.2)^k (0.8)^{20-k} \approx 0 \]Conclusion: Impossible to pass by random guessing.