Chapter 12.6, Problem 54SR

To determine

To find: Probability that there will be atleast $7$ successes.

Answer to Problem 54SR

Probability that there will be at least $7$ successes is about $0.6496$ .

Explanation of Solution

Given information:

Rate of success $=70\%$

It means

Probability of success $=0.7$

Total number of trails $=n=10$

Concept:

Formula of Binomial distribution is

  $b\left(x;n,P\right)=C{}_x\times P^x\times {\left(1-P\right)}^{n-x}$

Where

  $b=$ Binomial probability

  $x=$ total number of “successes”.

  $P=$ probability of a success on an individual trial.

  $n=$ Number of trials.

Given

  $\begin{array}{l}n=10\\ P=0.7\end{array}$

At least $7$ successes.

It means

Number of success $=7$ or

  $8$ or $9$ or $10$

Therefore,

Probability for at least $7$ successes is

  $\begin{array}{l}b=C{}_7{\left(0.7\right)}^7{\left(0.3\right)}^{10-7}+C{}_8{\left(0.7\right)}^8{\left(0.3\right)}^{10-8}+C{}_9{\left(0.7\right)}^9{\left(0.3\right)}^{10-9}+C{}_{10}{\left(0.7\right)}^{10}{\left(0.3\right)}^{10-10}\\ \Rightarrow b=\frac{10!}{\left(10-7\right)!7!}{\left(0.7\right)}^7{\left(0.3\right)}^3+\frac{10!}{\left(10-8\right)!8!}{\left(0.7\right)}^8{\left(0.3\right)}^2+\frac{10!}{\left(10-9\right)!9!}{\left(0.7\right)}^9{\left(0.3\right)}^1+\frac{10!}{\left(10-10\right)!10!}{\left(0.7\right)}^{10}{\left(0.3\right)}^0\\ \Rightarrow b=\frac{10\times 9\times 8\times 7!}{3!7!}{\left(0.7\right)}^7{\left(0.3\right)}^3+\frac{10\times 9\times 8!}{2!8!}{\left(0.7\right)}^8{\left(0.3\right)}^2+\frac{10\times 9!}{9!}{\left(0.7\right)}^9{\left(0.3\right)}^1+{\left(0.7\right)}^{10}\\ \Rightarrow b=120{\left(0.7\right)}^7{\left(0.3\right)}^3+45{\left(0.7\right)}^8{\left(0.3\right)}^2+10{\left(0.7\right)}^9{\left(0.3\right)}^1+{\left(0.7\right)}^{10}\\ \Rightarrow b=0.266827932+0.233474442+0.121060821+0.0282475249\\ \Rightarrow b=0.6496\end{array}$

Therefore,

Probability that there will be at least $7$ successes is about $0.6496$ .