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

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).

Explanation:

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})