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.


Expert's answer

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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