Answer:
a) [tex]P(X \leq 3) = 0.99328[/tex]
b) 0.6957
Step-by-step explanation:
Let X represent the number of 4's when n = 5 independent spins
each has a probability of 0.2 (i.e p = 0.2)
This notation is represented as:
X [tex]\approx[/tex] Binomial (n = 5, p = 0.2)
Probability of [tex]x[/tex] number of 4's is:
[tex]P(X=x)= (\left \ n \atop x \right) p^x (1-p)^{(n-x)}[/tex]
here; [tex](\left \ n \atop x \right)[/tex] is the combinatorial expression
[tex](\left \ n \atop x \right)[/tex] = [tex]\frac{n!}{x!(n-x)!}[/tex]
[tex]P(X \leq3), n =5 , p = 0.2[/tex]
[tex]P(X \leq3) = 1-P(X > 3)[/tex]
So; let's first find:
[tex]P(X > 3)[/tex]
[tex]= P(3 <X \leq 5) \\ \\ = P(4 <X \leq 5) \\ \\ = P (X = 4, 5) \\ \\ = P (X=4)+P(X = 5 ) \ \ \ (disjoint \ events)[/tex]
[tex]P(X = 4) =( \left \ {{5} \atop {4}} \right. ) (0.2)^4 (1-0.2)^1 \\ \\ P(X = 4) = 5(0.2)^4(0.8)^1 \\ \\ P(X = 4) = 0.0064[/tex]
[tex]P(X = 5) =( \left \ {{5} \atop {5}} \right. ) (0.2)^5 (1-0.2)^0 \\ \\ P(X = 5) = 5(0.2)^5(0.8)^0 \\ \\ P(X = 5) = 0.00032[/tex]
[tex]P (X=4)+P(X = 5 ) \\ \\ = 0.0064 + 0.00032 = 0.006720 \\ \\ \approx 0.007[/tex]
[tex]P(X > 3 ) = 0.00672 \\ \\ P(X \leq 3) = 1- P(X > = 3 ) \\ \\ =1 - 0.00672 \\ \\ = 0.99328[/tex]
[tex]P(X \leq 3) = 0.99328[/tex]
b)
Given that:
The ratio of boys to girls at birth in Singapore is quite high at 1.09:1
What proportion of Singapore families with exactly 6 children will have at least 3 boys?
Probability of having a boy = [tex]\frac{1.09}{1+1.09}[/tex] = 0.5215
Binomial Problem with n = 6
P(3<= x <=6) = 1 - P(0<= x <=2)
= 1 - binomial (6,0.5215,2)
= 0.6957