Question

Four coins are tossed. Let Z be the random variable representing the number of tails.

Answer 1

Solution:

We assume that the probabilities of getting heads and tails are the same

$p = q = \frac{1}{2}$

Using the Bernoulli formula, we find the probabilities that there will be 0, 1, 2, 3 and 4 tails, respectively

$P\left( {z = 0} \right) = {q^4} = \frac{1}{{16}}$

$P(z = 1) = C_4^1p{q^3} = \frac{4}{{16}} = \frac{1}{4}$

$P(z = 2) = C_4^2{p^2}{q^2} = \frac{6}{{16}} = \frac{3}{8}$

$P(z = 3) = C_4^3{p^3}q = \frac{4}{{16}} = \frac{1}{4}$

$P\left( {z = 4} \right) = {p^4} = \frac{1}{{16}}$

We have a distribution series

We assume that the probabilities of getting heads and tails are the same

$p = q = \frac{1}{2}$

Using the Bernoulli formula, we find the probabilities that there will be 0, 1, 2, 3 and 4 heads, respectively

$P\left( {z = 0} \right) = {q^4} = \frac{1}{{16}}$

$P(z = 1) = C_4^1p{q^3} = \frac{4}{{16}} = \frac{1}{4}$

$P(z = 2) = C_4^2{p^2}{q^2} = \frac{6}{{16}} = \frac{3}{8}$

$P(z = 3) = C_4^3{p^3}q = \frac{4}{{16}} = \frac{1}{4}$

$P\left( {z = 4} \right) = {p^4} = \frac{1}{{16}}$

We have a distribution series

We assume that the probabilities of getting heads and tails are the same

$p = q = \frac{1}{2}$

Using the Bernoulli formula, we find the probabilities that there will be 0, 1, 2, 3 and 4 heads, respectively

$P\left( {z = 0} \right) = {q^4} = \frac{1}{{16}}$

$P(z = 1) = C_4^1p{q^3} = \frac{4}{{16}} = \frac{1}{4}$

$P(z = 2) = C_4^2{p^2}{q^2} = \frac{6}{{16}} = \frac{3}{8}$

$P(z = 3) = C_4^3{p^3}q = \frac{4}{{16}} = \frac{1}{4}$

$P\left( {z = 4} \right) = {p^4} = \frac{1}{{16}}$

We have a distribution series