Chapter 12, Problem 39SGR

To determine

The probability of the given statement for no more than the 3 success.

Answer to Problem 39SGR

The probability is 38.23%.

Explanation of Solution

Given:

The binomial distribution has 40% rate of success, there are 10 trials.

Considering x as the random variable which shows the number of successes.

Rate of the success (s)=40%

Number of trials(n)=10

The binomial distribution is given as:

  $P\left(X=x\right)=C\left(n,x\right){\left(s\right)}^x{\left(f\right)}^{n-x}$

Here, s is the probability of the success of the trail and f =1-s

Substituting the known values into the above equation:

  $\begin{array}{l}P\left(X=x\right)=C\left(10,x\right){\left(0.4\right)}^x{\left(0.6\right)}^{10-x}\\ =C\left(10,x\right){\left(0.4\right)}^x{\left(0.6\right)}^{10-x}\end{array}$

Now, substitute the value of the x as 0,1,2, ……..,10 in the above equation:

The probability for the different value of the x will be:

  $\begin{array}{l}x=0\to P\left(X=x\right)=0.00604662\\ x=1\to P\left(X=x\right)=0.04031078\\ x=2\to P\left(X=x\right)=0.12093235\\ x=3\to P\left(X=x\right)=0.21499085\\ x=4\to P\left(X=x\right)=0.25082266\\ x=5\to P\left(X=x\right)=0.20065812\\ x=6\to P\left(X=x\right)=0.11147674\\ x=7\to P\left(X=x\right)=0.04246733\\ x=8\to P\left(X=x\right)=0.01061683\\ x=9\to P\left(X=x\right)=0.00157286\\ x=10\to P\left(X=x\right)=0.00010486\end{array}$

Now, evaluating the probability of no more than 3 successes:

  $\begin{array}{c}P\left(X\le 3\right)={\displaystyle \sum _{x=4}^{3}C\left(10,x\right)}{\left(0.4\right)}^x{\left(0.6\right)}^{10-x}\\ =0.00604662+0.04031078+0.12093235+0.21499085\\ =0.3823\\ =38.23\%\end{array}$

Therefore, the probability is 38.23%.