Question

1. jessica flips a coin 5 times. what is the probability that the coin lands on heads 2 times or fewer?

Explanation:

Step1: Identify the distribution

This is a binomial probability problem with \( n = 5 \) (number of trials), \( p=\frac{1}{2} \) (probability of success, heads), and we need \( P(X\leq2) \) where \( X \) is the number of heads. The binomial probability formula is \( P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k} \), and \( C(n,k)=\frac{n!}{k!(n - k)!} \).

Step2: Calculate \( P(X = 0) \)

For \( k = 0 \):
\( C(5,0)=\frac{5!}{0!(5 - 0)!}=1 \)
\( P(X = 0)=1\times(\frac{1}{2})^{0}\times(\frac{1}{2})^{5}=(\frac{1}{2})^{5}=\frac{1}{32} \)

Step3: Calculate \( P(X = 1) \)

For \( k = 1 \):
\( C(5,1)=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=5 \)
\( P(X = 1)=5\times(\frac{1}{2})^{1}\times(\frac{1}{2})^{4}=5\times\frac{1}{32}=\frac{5}{32} \)

Step4: Calculate \( P(X = 2) \)

For \( k = 2 \):
\( C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4\times3!}{2\times1\times3!}=10 \)
\( P(X = 2)=10\times(\frac{1}{2})^{2}\times(\frac{1}{2})^{3}=10\times\frac{1}{32}=\frac{10}{32} \)

Step5: Sum the probabilities

\( P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)=\frac{1}{32}+\frac{5}{32}+\frac{10}{32}=\frac{1 + 5+10}{32}=\frac{16}{32}=\frac{1}{2} \)

Answer:

\(\frac{1}{2}\)