Respuesta :
Answer:
a) 10*0.51^3*0.49^2=approx 0.32
b)1-0.49^5= approx 0.97
c)1-0.51^5=0.965
d) approx 0.028 =P(all kids are girls) approx 0.035 =P(all kids are boys)
P(all kids are of the same sexboys or girls)=0.063
Step-by-step explanation:
a) The probability that exactly 3 of 5 kids are boys is
p(3 boys)=C5 3 *p^3*q^2
C5 3 = 5!/(3!*2!)= 4*5/2=10
p=0.51 is the probability a child is boy
q=1-p=1-0.51=0.49 is the probability a child is girl
P(3 boys of 5 kids)= 10*0.51^3*0.49*2
b) The event that at least 1 kid from five is boy is a combination of events
1 boy -4 girls, 2 boys-3 girls, 3 boys-2 girls, 4 boys-1 girl, 5 boys.
The event " all 5 kids are the girls" is the only event which is not the part of
event " at least 1 kid from five is boy"
So P(at least 1 kid from five is boy)=1 - P(all 5 kids are the girls)
P(all 5 kids are the girls)=0.49^5=approx 0.028
P(at least 1 kid from five is boy)=1-0.028=approx 0.97
c) Similarly like b) P(at least one of five kids is girl)= 1-P(all 5 kids are the boys)
P(all 5 kids are the boys)=0.51^5=approx 0.035
P(at least one of five kids is girl)= 1-0.035=approx=0.965
d) Probability all kids are of the same sex= probability that all kids are the boys+ probability all kids are the girls.
Using b) and c) P(all kids of the same sex)= 0.035+0.028=0.063
The probability that a family of five children has all children of the same gender is 0.06274.
What is Binomial distribution?
A common discrete distribution is used in statistics, as opposed to a continuous distribution is called a Binomial distribution. It is given by the formula,
[tex]P(x) = ^nC_x p^xq^{(n-x)}[/tex]
Where,
x is the number of successes needed,
n is the number of trials or sample size,
p is the probability of a single success, and
q is the probability of a single failure.
Given the probability of a child being a boy is 0.51, therefore, p= 0.51. Also, we can write the probability of the child not being a boy is 0.49(1-0.51), therefore, the probability of the child being a girl is 0.49 and q=0.49.
A.) The probability that a family of five children is exactly three boys can be written as
[tex]P(x) = ^nC_x p^xq^{(n-x)}\\\\P(x=3)= ^5C_3 \cdot (0.51)^3 \cdot (0.49)^{2}\\\\P(x=3) = 0.3184[/tex]
B.) The probability that a family of five children has at least one boy can be written as,
Probability of at least one boy = 1 - Probability of no boy
[tex]P(x) = ^nC_x p^xq^{(n-x)}\\\\P(x=0)= ^5C_0 \cdot (0.51)^0 \cdot (0.49)^{5}\\\\P(x=0) = 0.02824[/tex]
Now,
[tex]\text{Probability of at least one boy }= 1 - \text{Probability of no boy}\\\\P(x > 1) = 1-P(x=0)\\\\P(x > 1) = 1-0.02824 = 0.9717[/tex]
C.) The probability that a family of five children has at least one girl can be written as,
Probability of at least one girl= 1 - Probability of no girl
[tex]P(x) = ^nC_x p^xq^{(n-x)}\\\\P(x=5)= ^5C_5 \cdot (0.51)^5 \cdot (0.49)^{0}\\\\P(x=5) = 0.0345[/tex]
Now,
[tex]\text{Probability of at least one girl}= 1 - \text{Probability of no girl}\\\\P(x < 5) = 1-P(x=5)\\\\P(x > 1) = 1-0.0345= 0.9654[/tex]
D.) The probability that a family of five children has all children of the same gender can be written as,
Probability = Probability of all boys + Probability of all girls
[tex]P = P(x=0)+P(x=5)\\\\P = 0.02824+ 0.0345\\\\P= 0.06274[/tex]
Hence, The probability that a family of five children has all children of the same gender is 0.06274.
Learn more about Binomial Distribution:
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