Respuesta :
Probability of an event is the measure of its chance of occurrence. The probability that the family has 0,1 or 2 girls is 0.7504
How to find that a given condition can be modeled by binomial distribution?
Binomial distributions consists of n independent Bernoulli trials.
Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as
[tex]X \sim B(n,p)[/tex]
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
What is the addition rule of probability for two events?
For two events A and B, we have:
Probability that event A or B occurs = Probability that event A occurs + Probability that event B occurs - Probability that both the event A and B occur simultaneously.
This can be written symbolically as:
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
For three events, A, B and C:
[tex]P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)[/tex]
For the given case,
If the random variable X is such that:
X = number of girls born to a family which has two children.
Then, we can model this situation by Binomial distribution, where Bernoulli experiment is birth of child which either results in being girl(call it success), or boy( call it failure) (in this case, we assume that only two gender exist).
Since total there are 2 children, so n = 2
In that specified family, success in Bernoulli experiment = birth of girl, so probability of success = P(birth of girl) = 0.48 = p
and, failure in Bernoulli experiment = birth of boy, thus, probability of failure = P(birth of boy) = 0.52 = q = 1-p
Thus, X pertains binomial distribution (as it is count of successes), as:
[tex]X \sim B(n = 2,p = 0.48)[/tex]
Evaluating the needed probabilities:
- Probability that the specified family gets 0 girl child = P(X = 0)
Using the probability function for binomial distribution, we get:
[tex]P(X = 0) = \: ^2C_0(0.48)^0(0.52)^2 = 0.2704\\[/tex]
- Probability that the specified family gets 1 girl child = P(X = 0)
Using the probability function for binomial distribution, we get:
[tex]P(X = 0) = \: ^2C_1(0.48)^1(0.52)^1 = 0.2496\\[/tex]
- Probability that the specified family gets 2 girl child = P(X = 0)
Using the probability function for binomial distribution, we get:
[tex]P(X = 0) = \: ^2C_2(0.48)^2(0.52)^0 = 0.2304\\[/tex]
If we take three events as:
- A = the specified family gets 0 girl child
- B = the specified family gets 1 girl child
- C = the specified family gets 2 girl child
Then, their simultaneous occurrence isn't possible, so the probability of their intersection isn't possible.
The probability of event of getting 0, 1 or 2 girl child in that family is written as: [tex]P(A \cup B\cup C)[/tex]. Using the addition rule, and knowing that their intersection is having 0 chances, thus,
[tex]P(A \cup B \cup C) = P(A) + P(B) + P(C) =[/tex] [tex]0.2704 + 0.2496 + 0.2304 = 0.7504[/tex]
Thus, he probability that the family has 0,1 or 2 girls is 0.7504
Learn more about binomial distributions here:
https://brainly.com/question/13609688
