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2) Assume that male and female births are equally likely and that the birth of any child does
not affect the probability of the gender of any other children. Find
the probability of at
most three boys in ten births.

Respuesta :

PoumonNeedsHelp

Answer: 17.14%

Step-by-step explanation:

To find the probability of at most three boys in ten births, we can use the binomial probability formula. The probability of getting exactly \( k \) successes (boys in this case) in \( n \) trials (births in this case) with the probability of success \( p \) is given by:

In this case:

- n (number of births) = 10

- b (number of boys) can be 0, 1, 2, or 3

- p (probability of a boy) = \( \frac{1}{2} \) since male and female births are equally likely

Let's calculate the probability for each case and sum them up:

P (X ≤ 3) = P (X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Then, calculate each term:

P (X = 0) = 10/0 * 1/2^0 * 1/2^10

P(X = 1) = 10/1 * 1/2^1 * 1/2^9

P(X = 2) = 10/2 * 1/2^2 * 1/2^8

P(X = 3) =  10/3 * 1/2^3 * 1/2^7

Calculating that gets you approximately 17.14% (if it needs to be rounded, then 17%)

msm555

Answer:

0.171875

Step-by-step explanation:

To find the probability of at most three boys in ten births using the binomial distribution formula, we need to compute the probabilities for [tex] \displaystyle k = 0, 1, 2, [/tex] and [tex] \displaystyle 3 [/tex], and then sum them up.

The formula for the probability of [tex] \displaystyle k [/tex] successes in [tex] \displaystyle n [/tex] trials is:

[tex] \displaystyle P(X = k) = \binom{n}{k} \times p^k \times (1 - p)^{n - k} [/tex]

Given:

[tex] \displaystyle n = 10 [/tex] (number of births)

We know that:

- [tex] \displaystyle p = 1/2 = 0.5 [/tex] (probability of having a boy)

We can calculate the probabilities as follows:

For [tex] \displaystyle k = 0 [/tex]:

[tex] \displaystyle P(X = 0) = \binom{10}{0} \times (0.5)^0 \times (1 - 0.5)^{10} [/tex]

[tex] \displaystyle = 1 \times 1 \times (0.5)^{10} [/tex]

[tex] \displaystyle = 1 \times 1 \times 0.0009765625 [/tex]

[tex] \displaystyle = 0.0009765625 [/tex]

For [tex] \displaystyle k = 1 [/tex]:

[tex] \displaystyle P(X = 1) = \binom{10}{1} \times (0.5)^1 \times (1 - 0.5)^{9} [/tex]

[tex] \displaystyle = 10 \times 0.5 \times (0.5)^9 [/tex]

[tex] \displaystyle = 0.009765625 [/tex]

For [tex] \displaystyle k = 2 [/tex]:

[tex] \displaystyle P(X = 2) = \binom{10}{2} \times (0.5)^2 \times (1 - 0.5)^{8} [/tex]

[tex] \displaystyle = 45 \times 0.25 \times (0.5)^8 [/tex]

[tex] \displaystyle = 0.0439453125 [/tex]

For [tex] \displaystyle k = 3 [/tex]:

[tex] \displaystyle P(X = 3) = \binom{10}{3} \times (0.5)^3 \times (1 - 0.5)^{7} [/tex]

[tex] \displaystyle = 120 \times 0.125 \times (0.5)^7 [/tex]

[tex] \displaystyle = 0.1171875 [/tex]

Now, summing these probabilities:

[tex] \displaystyle P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) [/tex]

[tex] \displaystyle = 0.0009765625 + 0.009765625 + 0.0439453125 + 0.1171875 [/tex]

[tex] \displaystyle = 0.171875 [/tex]

So, the probability of having at most three boys in ten births is [tex] \displaystyle 0.171875 [/tex].