The success rate of corneal transplant surgery is 85%. The surgery is performed on six patients. (Adapted from St. Luke's Cataract & Laser Institute)

(a) Construct a binomial distribution.

(b) Graph the binomial distribution using a probability histogram. Then describe its shape.

(c) Find the mean, variance, and standard deviation of the probability distribution and interpret the results.

(d) Find the probability that the surgery is successful for exactly three patients. Is this an unusual event? Explain.

(e) Find the probability that the surgery is successful for fewer than four patients. Is this an unusual event? Explain.

#### Solution

The success rate of corneal transplant surgery is 85%, i.e.,$p=0.85$. The surgery is performed on six patients, i.e., $n =6$. Thus $X\sim B(6, 0.85)$.

The probability mass function of $X$ is

`$$ \begin{aligned} P(X=x) &= \binom{6}{x} (0.85)^x (1-0.85)^{6-x}, \\ & \qquad x=0,1,\cdots, 6. \end{aligned} $$`

(a) Construct a binomial distribution

The probability that $X$ is exactly $0$ is

`$$ \begin{aligned} P(X= 0) & =\binom{6}{0} (0.85)^{0} (1-0.85)^{6-0}\\ & = 0\\ \end{aligned} $$`

The probability that $X$ is exactly $1$ is

`$$ \begin{aligned} P(X= 1) & =\binom{6}{1} (0.85)^{1} (1-0.85)^{6-1}\\ & = 0.0004\\ \end{aligned} $$`

The probability that $X$ is exactly $2$ is

`$$ \begin{aligned} P(X= 2) & =\binom{6}{2} (0.85)^{2} (1-0.85)^{6-2}\\ & = 0.0055\\ \end{aligned} $$`

The probability that $X$ is exactly $3$ is

`$$ \begin{aligned} P(X= 3) & =\binom{6}{3} (0.85)^{3} (1-0.85)^{6-3}\\ & = 0.0415\\ \end{aligned} $$`

The probability that $X$ is exactly $4$ is

`$$ \begin{aligned} P(X= 4) & =\binom{6}{4} (0.85)^{4} (1-0.85)^{6-4}\\ & = 0.1762\\ \end{aligned} $$`

The probability that $X$ is exactly $5$ is

`$$ \begin{aligned} P(X= 5) & =\binom{6}{5} (0.85)^{5} (1-0.85)^{6-5}\\ & = 0.3993\\ \end{aligned} $$`

The probability that $X$ is exactly $6$ is

`$$ \begin{aligned} P(X= 6) & =\binom{6}{6} (0.85)^{6} (1-0.85)^{6-6}\\ & = 0.3771\\ \end{aligned} $$`

x | Px |
---|---|

0 | 0.0000 |

1 | 0.0004 |

2 | 0.0055 |

3 | 0.0415 |

4 | 0.1762 |

5 | 0.3993 |

6 | 0.3771 |

(b) Graph of Binomial Distribution

The shape of Binomial distribution with $n=6$ and $p=0.85$ is negatively skewed.

(c) The mean of Binomial Distribution is

$E(X) = n*p = 6\times 0.85 = 5.1$

The variance of Binomial distribution is

`$V(X) = n*p*(1-p) = 6\times 0.85\times (1- 0.85) = 0.765$`

The standard deviation of Binomial distribution is

$sd(X) = \sqrt{V(X)}= \sqrt{0.765} = 0.8746$

(d) The probability that $X$ is exactly $3$ is

`$$ \begin{aligned} P(X= 3) & =\binom{6}{3} (0.85)^{3} (1-0.85)^{6-3}\\ & = 0.0415\\ \end{aligned} $$`

The probability of $X= 3$ is $0.0415$ which is very small. So the value of $X=3$ is unusual.

(e) The probability that $X$ is fewer than $4$ is

`$$ \begin{aligned} P(X< 4)& = P(X=0) + P(X=1)+P(X=2)+P(X=3)\\ & = 0 + 0.0004+ 0.0055+0.0415\\ & = 0.0473 \end{aligned} $$`

The probability of $X< 4$ is $0.0473$ which is very small. So the event of fewer than four patients is unusual event.