Question

An urn contains 6 red balls, 8 green balls, and 10 white balls. If balls are selected one by one (without replacing them after they are selected) what is the probability that at least two balls must be selected in order to get a green ball. g

Answer 1

Answer:

0.43478

Step-by-step explanation:

From the information given:

$The \ required \ probability = P(1st ball = red) \times P(2nd ball = red) + P(1st ball = red) \times P(2nd ball = white) + P(2nd ball = red) \times P(1st ball = white) + P(1st ball = white) \times P(2nd ball = white)$

$= \dfrac{6}{6+8+10}\times \dfrac{5}{5+8+10} + \dfrac{6}{6+8+10} \times \dfrac{10}{5+8+10} + \dfrac{10}{6+8+10} \times \dfrac{6}{5+8+10} \times \dfrac{10}{6+8+10} \times \dfrac{9}{5+8+9}$

$=(\dfrac{6}{24}\times \dfrac{5}{23})+ (\dfrac{6}{24}\times \dfrac{10}{23}) +( \dfrac{10}{24}\times \dfrac{6}{23} ) + (\dfrac{10}{24}\times \dfrac{9}{23})$

= 0.05434782609 + 0.1086956522  + 0.1086956522  + 0.1630434783

= 0.4347826088

≅ 0.43478