Answer to Question #340158 in Statistics and Probability for Maneo

Question

1.Let the random variable X be the total number of heads when a fair coin is tossed 4 times.

•Is this a Bernoulli or binomial trial/experiment?

•Are the conditions of a Binomial trial satisfied?

•Find the probability mass function for X.

2. The probability that the patient recovers from a rare blood disease is 0.4. If 10 people are known to have contracted the disease, what is the probability that

a) exactly 3 patients survive the disease?

b) from 3 to 5 patients survive the disease?

3. Suppose that a student is given a test with 10 true-false questions. Let X be the number of questions guessed correctly by the student.

i) Write down the probability distribution (probability mass function) of X

ii) What is the probability that the student guesses all questions correctly?

iii) what is the probability that is the probability that the student guesses 7 or more questions

correctly?

Answer 1

1.

This a Bernoulli or binomial trial/experiment.

The conditions of a Binomial trial are satisfied

1) There are a fixed number of trials (a fixed sample size).

2) On each trial, the event of interest either occurs or does not.

3) The probability of occurrence (or not) is the same on each trial.

4) Trials are independent of one another.

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

$$n=4, p=0.5, q=1-p=0.5$$

$$P(X=x)=\dbinom{4}{x}(0.5)^{x}(0.5)^{4-x}, x=0,1,2,3,4$$

2.

$$n=10, p=0.4, q=1-p=0.6$$

i)

$$P(X=3)=\dbinom{10}{3}(0.4)^{3}(0.6)^{10-3}=0.214990848$$

ii)

$$P(3\le X\le 5)=P(X=3)+P(X=4)+P(X=5)$$

Of

$$=\dbinom{10}{3}(0.4)^{3}(0.6)^{10-3}+\dbinom{10}{4}(0.4)^{4}(0.6)^{10-4}$$

$$+\dbinom{10}{5}(0.4)^{5}(0.6)^{10-5}=0.214990848$$

$$+0.250822656+0.2006581248$$

$$=0.6664716288$$

3.

$$n=10, p=0.5, q=1-p=0.5$$

(i)

$$P(X=x)=\dbinom{10}{x}(0.5)^{x}(0.5)^{10-x},$$

$$x=0,1,2,3,4,5,6,7,8,9,10$$

(ii)

$$P(X=10)=\dbinom{10}{10}(0.5)^{10}(0.5)^{10-10}$$

$$=0.0009765625$$

(iii)

$$P( X\ge 7)=P(X=7)+P(X=8)+P(X=9)$$

$$+P(X=10)=\dbinom{10}{7}(0.5)^{7}(0.5)^{10-7}$$

$$+\dbinom{10}{8}(0.5)^{8}(0.5)^{10-8}+\dbinom{10}{9}(0.5)^{9}(0.5)^{10-9}$$

$$+\dbinom{10}{10}(0.5)^{10}(0.5)^{10-10}=0.171875$$