Question

Three red balls, 4 white balls, and 1 green ball are in a bag. A ball is drawn without replacement. What is the probability that 3 balls can be drawn without drawing the green ball?

Answer 1

Answer:

$\frac{5}{8}$

Step-by-step explanation:

Three red balls, 4 white balls, and 1 green ball are in a bag. A ball is drawn without replacement.

total balls = 3 red + 4 white + 1 green = 8 balls

the number of balls that are not green is 7

pick 3 balls from 7 balls that are not green

$nCr= \frac{n!}{r!(n-r)!}$

$P(not \ green)=\frac{7C3}{8C3}$

$7C3= \frac{7!}{3!(7-3)!}=35$

$8C3= \frac{8!}{3!(8-3)!} =56$

$P(3 \ balls \ drawn)=\frac{35}{56} =\frac{5}{8}$