Question

math hw 2

The Binomial Distribution



pls help me with my homework!!


thanks,

Answer 1

Using the binomial distribution, we have that:

a) The distribution is given by:

• P(X = 0) = 0.0625.

• P(X = 1) = 0.25.

• P(X = 2) = 0.375.

• P(X = 3) = 0.25.

• P(X = 4) = 0.0625.

b) 0.3125 = 31.25%.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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.

In this problem, we have that:

• There are 4 children, hence n = 4.

• Each children has a 50% chance of being female, hence p = 0.5.

The distribution is given by the probability of each outcome, hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.5)^{0}.(0.5)^{4} = 0.0625$

$P(X = 1) = C_{4,1}.(0.5)^{1}.(0.5)^{3} = 0.25$

$P(X = 2) = C_{4,2}.(0.5)^{2}.(0.5)^{2} = 0.375$

$P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25$

$P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625$

The probability of at least 3 daughters is given by:

$P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125$.

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

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