Question

f) find the probability of having two or less heads.

Explanation:

Assume we are flipping a fair coin \(n\) times. Let's assume \(n = 3\) (since the problem doesn't specify the number of coin - flips, and 3 is a common number for such problems). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success on a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n\) is the number of coin - flips, \(k\) is the number of heads, and \(p=\frac{1}{2}\) (probability of getting a head on a single coin - flip).

Step1: Calculate \(P(X = 0)\)

For \(n = 3\), \(k = 0\), \(p=\frac{1}{2}\), \(C(3,0)=\frac{3!}{0!(3 - 0)!}=1\), \(P(X = 0)=C(3,0)\times(\frac{1}{2})^{0}\times(1-\frac{1}{2})^{3 - 0}=1\times1\times(\frac{1}{2})^{3}=\frac{1}{8}\)

Step2: Calculate \(P(X = 1)\)

For \(n = 3\), \(k = 1\), \(C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\), \(P(X = 1)=C(3,1)\times(\frac{1}{2})^{1}\times(1 - \frac{1}{2})^{3 - 1}=3\times\frac{1}{2}\times(\frac{1}{2})^{2}=\frac{3}{8}\)

Step3: Calculate \(P(X = 2)\)

For \(n = 3\), \(k = 2\), \(C(3,2)=\frac{3!}{2!(3 - 2)!}=3\), \(P(X = 2)=C(3,2)\times(\frac{1}{2})^{2}\times(1-\frac{1}{2})^{3 - 2}=3\times(\frac{1}{2})^{2}\times\frac{1}{2}=\frac{3}{8}\)

Step4: Calculate \(P(X\leq2)\)

\(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)=\frac{1}{8}+\frac{3}{8}+\frac{3}{8}=\frac{7}{8}\)

Answer:

\(\frac{7}{8}\)