Answer to Question #149145 in Statistics and Probability for Balwinder Kaur

Question

8. A box contains 22 balls of which 7 are red, 9 are white, and 6 are black. Four balls are drawn at random from the box without replacement. Find the probability that

(a) three red balls and one white ball are drawn,

(b) the four balls are all red,

(c) the balls are all of the same colour,

(d) there is at least one ball of each colour.

Answer 1

(a) Total amount of combinations C(22, 4) = 7315.

$\displaystyle P(A) = \frac{C(7,3) \cdot C(9,1)}{C(22,4)} = \frac{35 \cdot 9}{7315} = 0.043$

(b)

$\displaystyle P(B) = \frac{C(7,4)}{C(22,4)} = \frac{25}{7315} = 0.0034$

(c) P(C) = P(4 red) + P(4 white) + P( 4 black)

$\displaystyle P(C) = 0.0034 + \frac{C(9,4)}{C(22,4)} + \frac{C(6,4)}{C(22,4)} = 0.0034 + \frac{126}{7315}+\frac{15}{7315} = 0.0034 + 0.0172 + 0.0020 = 0.0226$

(d) P(D) = P(2r 1w 1b) + P(1r 2w 1b) + P(1r 1w 2b)

$\displaystyle P(D) = \frac{C(7,2) \cdot C(9,1) \cdot C(6,1)}{C(22,7)} + \frac{C(7,1) \cdot C(9,2) \cdot C(6,1)}{C(22,7)} + \frac{C(7,1) \cdot C(9,1) \cdot C(6,2)}{C(22,7)} = \frac{21 \cdot 9 \cdot6}{7315} + \frac{7 \cdot 36 \cdot 6}{7315} + \frac{7 \cdot 9 \cdot 15}{7315} = 0.155+ 0.207+0.129 = 0.491$