3)A hockey coach has 5 white corks,4 orange corks and some luminous green corks in the bag for drills during practice.The balls are similar in shape and size;
a)If the probability of picking a luminous green ball is 0.25,find the total number of balls in the bag
b)Three corks picked at random,one after the other without replacement and their colour noted.Find the probability that the three balls are of:(i)The same colour
(ii)Different colour
c)The probability of Jack winning a game of pool is 2/3.If he plays the game until he wins,find his probability of winning the game in either the third or fourth attempt?
a) Let the total number of balls in the bag be x.
The probability of picking a luminous green ball is given as 0.25, so the number of luminous green balls is 0.25x.
We know that the total number of white balls is 5, and the total number of orange balls is 4.
Since the total number of balls in the bag is x, we can write the equation:
0.25x + 5 + 4 = x
0.25x - x = -9
-0.75x = -9
x = -9 / -0.75
x = 12
Therefore, the total number of balls in the bag is 12.
b) (i) To find the probability that the three balls are of the same color, we need to calculate the probability of picking three white balls or three orange balls.
P(Three white balls) = (5/12) * (4/11) * (3/10) = 60/1320 = 1/22
P(Three orange balls) = (4/12) * (3/11) * (2/10) = 24/1320 = 1/55
P(Three balls of the same color) = P(Three white balls) + P(Three orange balls) = 1/22 + 1/55 = 3/110
(ii) To find the probability that the three balls are of different colors, we need to calculate the probability of picking one white ball, one orange ball, and one luminous green ball.
P(One white ball, one orange ball, one luminous green ball) = (5/12) * (4/11) * (0.25) = 10/1320 = 1/132
c) The probability of winning a game of pool for Jack is 2/3. This means that the probability of losing a game is 1/3.
The probability of Jack winning the game in his third attempt is the probability of losing the first two attempts (1/3) and winning on the third attempt (2/3).
P(Jack winning in the third attempt) = (1/3) * (1/3) * (2/3) = 2/27
The probability of winning the game in the fourth attempt is the probability of losing the first three attempts (1/3) and winning on the fourth attempt (2/3).
P(Jack winning in the fourth attempt) = (1/3) * (1/3) * (1/3) * (2/3) = 2/81
Therefore, the probability of Jack winning the game in either the third or fourth attempt is:
P(Jack winning in the third or fourth attempt) = P(Jack winning in the third attempt) + P(Jack winning in the fourth attempt) = 2/27 + 2/81 = 8/81