Question

Find each of the following probabilities when n independent Bernoulli trials are carried out with a probability of success p.
a. the probability of no successes.
b. the probability of at least one success.
c. the probability of at most one success.
d. the probability of at least two successes.

Answer 1

Answer:

a)$P(X=0) = (nC0)(p)^0 (1-p)^{n-0}=1*1* (1-p)^n =(1-p)^n$

b) $P(X\geq 1)=1-P(X<1)= 1-P(X=0)= 1-(1-p)^n$

c) $P(x\leq 1) = (1-p)^n+ (nC1)(p)^1 (1-p)^{n-1} =(1-p)^n +np(1-p)^{n-1}$

d) $P(x \leq 2)= 1-[P(x=0)+P(X=1)]=1-[(1-p)^n + np(1-p)^{n-1}]$

Step-by-step explanation:

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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:

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

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

a. the probability of no successes.

$P(X=0) = (nC0)(p)^0 (1-p)^{n-0}=1*1* (1-p)^n =(1-p)^n$

b. the probability of at least one success.

On this case we are looking for this probability, and we can use the complement rule and the result from part a to solve it:

$P(X\geq 1)=1-P(X<1)= 1-P(X=0)= 1-(1-p)^n$

c. the probability of at most one success.

On this case we are looking for this probability:

$P(x\leq 1)=P(X=0)+ P(X=1)$

And replacing the rrespective formulas we have:

$P(x\leq 1) = (1-p)^n+ (nC1)(p)^1 (1-p)^{n-1} =(1-p)^n +np(1-p)^{n-1}$

d. the probability of at least two successes.

On this case we are looking for this probability, and we can use the complement rule and the result from part a to solve it:

$P(x \geq 2) = 1-P(X<2) = 1- P(x\leq 1)$

And this is equivalent to:

$P(x \leq 2)= 1-[P(x=0)+P(X=1)]=1-[(1-p)^n + np(1-p)^{n-1}]$

Answer 2

The probability expressions are

• P(0) = (1 - p)^n
• P(x ≥ 1) = 1 - (1 - p)^n
• $P(x \le 1) = (1 - p)^n + np(1 - p)^{n-1}$
• $P(x \ge 2) = 1 - (1 - p)^n - np(1 - p)^{n-1$

How to determine the probabilities?

The given parameters are:

• Sample size = n
• Probability of success = n

The probability of a Bernoulli trial is calculated using:

$P(x) = ^nC_x *p^x * (1 - p)^{n-x$

a. the probability of no successes.

This means that x = 0;

So, we have:

$P(0) = ^nC_0 *p^0 * (1 - p)^n$

This gives

$P(0) = 1 *1 * (1 - p)^n$

P(0) = (1 - p)^n

b. the probability of at least one success.

This means that x = 1, 2,3....

To do this, we use the following complement rule

P(x ≥ 1) = 1 - P(0)

So, we have:

P(x ≥ 1) = 1 - (1 - p)^n

c. the probability of at most one success.

This means that:

x = 0, 1.

So, we have:

P(x ≤ 1) = P(0) + P(1)

Calculate P(1)

$P(1) = ^nC_1 *p^1 * (1 - p)^{n-1}$

$P(1) = np(1 - p)^{n-1}$

So, we have:

$P(x \le 1) = (1 - p)^n + np(1 - p)^{n-1}$

d. the probability of at least two successes.

This means that x = 2,3....

To do this, we use the following complement rule

P(x ≥ 2) = 1 - P(x ≤ 1)

So, we have:

$P(x \ge 2) = 1 - (1 - p)^n - np(1 - p)^{n-1$

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