Chapter 13.6, Problem 21E

To determine

To find: the probability of getting a $4$ heads if the coin is tossed $4$ times.

Answer to Problem 21E

The probability of getting $4$ heads is $\frac{1}{81}$ .

Explanation of Solution

Given:

The probability of tossing a head on a bent coin is $\frac{1}{3}$ .

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:

According to the given the probability of tossing a head on a bent coin is $\frac{1}{3}$ .

Getting $4$ heads means each of the $4$ flips is heads. The probability of getting heads once is $\frac{1}{3}$ ,so by Multiply by $\frac{1}{3}$ for each time tossed gets heads.

  $P\left(4\text{ heads}\right)=\left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{1}{81}.$

Hence, the probability of getting $4$ headsis $\frac{1}{81}$ .