Answer to Question #216999 in Statistics and Probability for Matie

Question

A bag contains 3 red, 7 black, 5 green marbles. Four marbles are selected at random without replacement. What is the probability that all 4 marbles are black?

Find the probability that no more than 2 green marbles are chosen? What is the probability that exactly 2 marbles are green and none are red?

Answer 1

Solution:

Red marbles=3

Black marbles=7

Green marbles=5

(a) Required probability to get all 4 black marbles$=\dfrac{^7C_4}{^{15}C_4}=\dfrac1{39}$

(b) Required probability to get no more than 2 green marbles = 0 green marble + 1 green marble + 2 green marbles$=\dfrac{^5C_0\times^{10}C_4}{^{15}C_4}+\dfrac{^5C_1\times^{10}C_3}{^{15}C_4}+\dfrac{^5C_2\times^{10}C_2}{^{15}C_4}$

$=\dfrac{12}{13}$

(c) Required probability to get exactly 2 green marbles and none are red

$=\dfrac{^5C_2\times^{7}C_2}{^{15}C_4} \\=\dfrac{2}{13}$