Question

The dingram shows a spinner innde up of a picce of card in the shape of a regular pentagon, with a toothpick pushed through its ceuter. The five triangles are mambered from I to 5. Each time. the spner is spin atil it lands on one of the five edges of the pentagon. The spinener is spun five tinmes. Use the binomial probability formula to enleulate the probability of at most three 4'sThe ratio of boys to girts 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? (ignore the probability of multiple births)​ what is the answer?

Answer 1

Answer:

a) $P(X \leq 3) = 0.99328$

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 $\approx$ Binomial (n = 5, p = 0.2)

Probability of  $x$  number of 4's is:

$P(X=x)= (\left \ n \atop x \right) p^x (1-p)^{(n-x)}$

here; $(\left \ n \atop x \right)$ is the combinatorial expression

$(\left \ n \atop x \right)$ = $\frac{n!}{x!(n-x)!}$

$P(X \leq3), n =5 , p = 0.2$

$P(X \leq3) = 1-P(X > 3)$

So; let's first find:

$P(X > 3)$

$= P(3$

$P(X = 4) =( \left \ {{5} \atop {4}} \right. ) (0.2)^4 (1-0.2)^1 \\ \\ P(X = 4) = 5(0.2)^4(0.8)^1 \\ \\ P(X = 4) = 0.0064$

$P(X = 5) =( \left \ {{5} \atop {5}} \right. ) (0.2)^5 (1-0.2)^0 \\ \\ P(X = 5) = 5(0.2)^5(0.8)^0 \\ \\ P(X = 5) = 0.00032$

$P (X=4)+P(X = 5 ) \\ \\ = 0.0064 + 0.00032 = 0.006720 \\ \\ \approx 0.007$

$P(X > 3 ) = 0.00672 \\ \\ P(X \leq 3) = 1- P(X > = 3 ) \\ \\ =1 - 0.00672 \\ \\ = 0.99328$

$P(X \leq 3) = 0.99328$

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 = $\frac{1.09}{1+1.09}$ = 0.5215

Binomial Problem with n = 6

P(3<= x <=6) = 1 - P(0<= x <=2)

= 1 - binomial (6,0.5215,2)

= 0.6957