Chapter 13.6, Problem 32E

To determine

To find: the probability of getting at least $2$ heads from $3$ coins tossed.

Answer to Problem 32E

The probability of getting at least $2$ heads is $\frac{1}{2}$ .

Explanation of Solution

Given:

  $3$ coins are tossed.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

The probability of getting head or tail from one coin is:

  $\begin{array}{l}P\left(\text{head}\right)=\frac{1}{2}.\\ P\left(\text{tail}\right)=\frac{1}{2}.\end{array}$

Here according to the given $3$ coins are tossed the probability of getting at least $2$ heads is:

  $\begin{array}{l}P\left(\text{at least }2\text{ heads }\right)=P\left(2\text{ heads}\right)+P\left(3\text{ heads}\right).\\ \Rightarrow C{}_2\times {\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^1+C{}_3\times {\left(\frac{1}{2}\right)}^3\times {\left(\frac{1}{2}\right)}^0.\\ \Rightarrow 3\times \frac{1}{4}\times \frac{1}{2}+1\times \frac{1}{8}\times 1.\\ \Rightarrow \frac{4}{8}=\frac{1}{2}.\end{array}$

Hence, the probability of gettingat least $2$ heads is $\frac{1}{2}$ .