Question

Each of urns I and II has 5 red balls, 3 white balls, and 2 green balls. Urn III has 1 red ball, 1 white ball, and 8 green balls. An urn is selected at random and a ball is drawn. If the ball is green, find the probability that urn III was selected. (Enter your probabilities as fractions.)

Answer 1

Answer:

The probability of selecting the green ball from urn III is ²/₃

Step-by-step explanation:

Given;

total number of balls in urn I = 10

number of green balls in urn I , = 2

total number of balls in urn II = 10

number of green balls in urn II , = 2

total number of balls in urn III = 10

number of green balls in urn II , = 8

$P = \frac{\frac{8}{10} }{\frac{2}{10} + \frac{2}{10}+ \frac{8}{10}}\\\\P = \frac{\frac{8}{10} }{\frac{12}{10} }\\\\P = \frac{8}{10} *\frac{10}{12} \\\\P = \frac{8}{12}\\\\ P = \frac{2}{3}$

Therefore, the probability of selecting the green ball from urn III is ²/₃