Question #169979

Five coins are tossed. Let Z be the random variable representing the number of heads that occur. Find the values of the random variable Z.


1
Expert's answer
2021-03-09T13:20:12-0500

We will assume that the probability of getting heads and tails is the same: p=q=12p = q = \frac{1}{2}.

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

P(0)=q5=(12)5=132P(0) = {q^5} = {\left( {\frac{1}{2}} \right)^5} = \frac{1}{{32}}


P(1)=C51pq4=5(12)5=532P(1) = C_5^1p{q^4} = 5 \cdot {\left( {\frac{1}{2}} \right)^5} = \frac{5}{{32}}


P(2)=C52p2q3=10(12)5=1032P(2) = C_5^2{p^2}{q^3} = 10 \cdot {\left( {\frac{1}{2}} \right)^5} = \frac{{10}}{{32}}


P(3)=C53p3q2=10(12)5=1032P(3) = C_5^3{p^3}{q^2} = 10 \cdot {\left( {\frac{1}{2}} \right)^5} = \frac{{10}}{{32}}


P(4)=C54p4q=5(12)5=532P(4) = C_5^4{p^4}q = 5 \cdot {\left( {\frac{1}{2}} \right)^5} = \frac{5}{{32}}


P(5)=p5=(12)5=132P(5) = {p^5} = {\left( {\frac{1}{2}} \right)^5} = \frac{1}{{32}}


We get the distribution law

Z012345p13253210321032532132\begin{matrix} Z&0&1&2&3&4&5\\ p&{\frac{1}{{32}}}&{\frac{5}{{32}}}&{\frac{{10}}{{32}}}&{\frac{{10}}{{32}}}&{\frac{5}{{32}}}&{\frac{1}{{32}}} \end{matrix}


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