Question

A new surgical procedure is successful with probability of p. Assume that the operation is performed 5 times and the results are independent of one another. Find the following probabilities. (You may use EXCEL if you want) What is the probability that all five operations are successful if p = 0.8? What is the probability that exactly four are successful is p = 0.6? What is the probability that less than two are successful is p = 0.3? What is the expected value of the number of successful operations when p = 0.5? What is the variance of the number of successful operations when p = 0.4?

Answer 1

Answer:

a) $P(X=5)=(5C5)(0.8)^5 (1-0.8)^{5-5}=0.32768$

And we can use the following excel code: "=BINOM.DIST(5,5,0.8,FALSE)"

b) $P(X=4)=(5C4)(0.6)^4 (1-0.6)^{5-4}=0.2592$

And we can use the following excel code: "=BINOM.DIST(4,5,0.6,FALSE)"

c) $P(X<2) = P(X \leq 1) = P(X=0) +P(X=1) =0.16807+0.36015=0.52822$

And the excel code would be: "=BINOM.DIST(1,5,0.3,TRUE)"

d) $E(X) =n*p= 5*0.5 =2.5$

e) $Var (X) = np(1-p) = 5*0.4*(1-0.4)=1.2$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=5, p)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case the value of p =0.8 and we want to find this probability:

$P(X=5)=(5C5)(0.8)^5 (1-0.8)^{5-5}=0.32768$

And we can use the following excel code: "=BINOM.DIST(5,5,0.8,FALSE)"

Part b

For this case the value of p =0.6 and we want this probability:

$P(X=4)=(5C4)(0.6)^4 (1-0.6)^{5-4}=0.2592$

And we can use the following excel code: "=BINOM.DIST(4,5,0.6,FALSE)"

Part c

The value of p assumed is 0.3. For this case we want this probability:

$P(X<2) = P(X \leq 1) = P(X=0) +P(X=1)$

$P(X=0)=(5C0)(0.3)^4 (1-0.3)^{5-0}=0.16807$

$P(X=1)=(5C1)(0.3)^4 (1-0.3)^{5-1}=0.36015$

$P(X<2) = P(X \leq 1) = P(X=0) +P(X=1) =0.16807+0.36015=0.52822$

And the excel code would be: "=BINOM.DIST(1,5,0.3,TRUE)"

Part d

The expected value for the binomial distribution is given by $np$ and since we are assuming p=0.5 we got:

$E(X) =n*p= 5*0.5 =2.5$

Part e

Assuming the value of p=0.4. The variance for the binomial distribution is given by:

$Var (X) = np(1-p) = 5*0.4*(1-0.4)=1.2$