44. Each child born to a particular set of parents has a probability of 0.25 of having blood type O. If these parents have 5 children.
What is the probability that
a. Exactly two of them have blood type O
b. At most 2 have blood type O
c. At least 4 have blood type O

Respuesta :

Using the binomial distribution, the probabilities are given as follows:

a. 0.2637 = 26.37%.

b. 0.8965 = 89.65%.

c. 0.0148 = 1.48%.

What is the binomial distribution formula?

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 5, p = 0.25.

Item a:

The probability is P(X = 2), hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637[/tex]

Item b:

The probability is:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

In which:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373[/tex]

[tex]P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955[/tex]

[tex]P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637[/tex]

Then:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2373 + 0.3955 + 0.2637 = 0.8965[/tex]

Item c:

The probability is:

[tex]P(X \geq 4) = P(X = 4) + P(X = 5)[/tex]

In which:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0147[/tex]

[tex]P(X = 5) = C_{5,1}.(0.25)^{5}.(0.75)^{0} = 0.0001[/tex]

Then:

[tex]P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0147 + 0.0001 = 0.0148[/tex]

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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