Question

A bowl has 6 red marbles and 3 green marbles. The probability that a blind folded person will draw a red marble on the second draw from the bowl without replacing the marble from the first draw, is

• A. $2/3$
• B. $1/4$
• C. $5/12$
• D. $5/8$

Answer 1

Answer: (A)

$2/3$

Given, in a bowl 6 red marbles and 3 green marbels.

Probability of getting 2nd draw as red$=$Probability of taking $first$ as $red$ and $second$ as $red$$+$probability of taking $first$ as $green$ and $second$ as $red$

$=(\frac{6}{9}\times \frac{5}{8})+(\frac{3}{9}\times \frac{6}{8})=\frac{2}{3}$