Answer:
(1) \(\frac{1}{9}\)
(2) \(\frac{8}{75}\)
(3) \(\frac{5}{12}\)
(4) \(\frac{3}{10}\)
Step-by-step explanation:
(1). A bag has 5 red marbles, 6 blue marbles and 4 black marbles.
Total marbles in a bag = 5 + 6 + 4 = 15 marbles
The probability of picking a red marble \(P_{1}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\)
By replacing it and then picking another red marble \(P_{2}\)= \(\frac{5}{15}\) = \(\frac{1}{3}\)
P = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)
(2). A bag has 5 red marbles, 6 blue marbles and 4 black marbles.
Total marbles in a bag = 5 + 6 + 4 = 15 marbles.
The probability of picking a blue marble = \(P_{1}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)
replacing it and then picking a black marble = \(P_{2}\) = \(\frac{4}{15}\)
P = \(\frac{2}{5}\) × \(\frac{4}{15}\) = \(\frac{8}{75}\)
(3). A bag contains 4 white, 3 black and 5 red marbles.
Total marbles = 4 + 3 + 5 = 12 marbles
The probability of picking a red marble = \(\frac{5}{12}\)
(4). A bag contains 3 white marbles, 2 black marbles and 5 red marbles.
Total marbles = 3 + 2 + 5 = 10 marbles.
The probability of picking a white marble = \(\frac{3}{10}\)
(1) \(\frac{1}{9}\)
(2) \(\frac{8}{75}\)
(3) \(\frac{5}{12}\)
(4) \(\frac{3}{10}\)
