Question

Let x be a random variable representing the number of tails which occur in tossing 5 coins

Answer 1

We will assume that the probability of getting heads and tails is the same: $p = q = \frac{1}{2}$.

Using Bernoulli's formula, we find the probability that 0, 1, 2, 3, 4, and 5 tails will land:

$$P(X=0)=\dbinom{5}{0}\big(\dfrac{1}{2}\big)^0 \big(\dfrac{1}{2}\big)^{5-0}=\dfrac{1}{32}$$

$$P(X=1)=\dbinom{5}{1}\big(\dfrac{1}{2}\big)^1 \big(\dfrac{1}{2}\big)^{5-1}=\dfrac{5}{32}$$

$$P(X=2)=\dbinom{5}{2}\big(\dfrac{1}{2}\big)^2 \big(\dfrac{1}{2}\big)^{5-2}=\dfrac{10}{32}$$

$$P(X=3)=\dbinom{5}{3}\big(\dfrac{1}{2}\big)^3 \big(\dfrac{1}{2}\big)^{5-3}=\dfrac{10}{32}$$

$$P(X=4)=\dbinom{5}{4}\big(\dfrac{1}{2}\big)^4 \big(\dfrac{1}{2}\big)^{5-4}=\dfrac{5}{32}$$

$$P(X=5)=\dbinom{5}{5}\big(\dfrac{1}{2}\big)^5 \big(\dfrac{1}{2}\big)^{5-5}=\dfrac{1}{32}$$

We get the distribution of the number of tails $X$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \\ p(x) & \dfrac{1}{32} & \dfrac{5}{32} & \dfrac{10}{32} & \dfrac{10}{32} & \dfrac{5}{32} & \dfrac{1}{32} \end{array}$$