Question

A box contains 10 green balls and 8 red balls. Six balls are selected at random. Find the probability that:

a) all 6 balls are green.

b) exactly 4 out of the 6 balls are red.

c) at least one of the 6 balls is red.

d) 3 or 4 of the 6 balls are red.

Answer 1

$$10+8=18(balls)$$

a)

$$P(6\ green)=\dfrac{\dbinom{10}{6}\dbinom{8}{0}}{\dbinom{18}{6}}=\dfrac{210(1)}{18564}=\dfrac{5}{442}$$

$$\approx0.011312$$

(b)

$$P(4\ red)=\dfrac{\dbinom{10}{2}\dbinom{8}{4}}{\dbinom{18}{6}}=\dfrac{45(70)}{18564}=\dfrac{75}{442}$$

$$\approx0.169683$$

(c)

$$P(at\ least\ 1\ red)=1-P(6\ green)$$

$$=1-\dfrac{5}{442}=\dfrac{437}{442}$$

$$\approx0.988682$$

(d)

$$P(3\ red\ or\ 4\ red)=P(3\ red)+P(4\ red)$$

$$=\dfrac{\dbinom{10}{3}\dbinom{8}{3}}{\dbinom{18}{6}}+\dfrac{\dbinom{10}{2}\dbinom{8}{4}}{\dbinom{18}{6}}$$

$$=\dfrac{120(56)}{18564}+\dfrac{45(70)}{18564}=\dfrac{235}{442}$$

$$\approx0.525727$$