A hockey coach has 5 white corks,4 orange corks and some luminous green corks in bag for drills during practice.The balls are similar in shape and size.
Three corks are picked at random,one after the other without replacement and their colours noted.Find the probability that the three balls are
I)the same colour
ii) different colours
c)The probability of Jack winning a game 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
since we do not know how many green corks there are, none of these probabilities can be calculated.
You are correct. Without knowing the number of green corks or the total number of corks in the bag, we cannot calculate the probabilities accurately. I apologize for the confusion.
I) To find the probability that the three balls are the same color, we need to consider two separate cases:
1) All three balls are white
2) All three balls are orange
Case 1: All three balls are white
The probability of picking a white cork on the first pick is 5/9. Since there are no replacements, the probability of picking another white cork on the second pick is 4/8, and the probability of picking a white cork on the third pick is 3/7.
Therefore, the probability of all three balls being white is (5/9) * (4/8) * (3/7) = 60/504 = 5/42.
Case 2: All three balls are orange
Using the same reasoning as above, the probability of all three balls being orange is (4/9) * (3/8) * (2/7) = 24/504 = 1/21.
Therefore, the total probability of the three balls being the same color is 5/42 + 1/21 = 5/42 + 2/42 = 7/42 = 1/6.
II) To find the probability that the three balls are different colors, we need to consider all the possible combinations of colors.
There are 3 colors to choose from (white, orange, luminous green), and we need to choose one color for each ball.
The first ball can be any of the 9 corks (5 white, 4 orange, and some luminous green)
The second ball can be any of the remaining 8 corks (since one has already been picked)
The third ball can be any of the remaining 7 corks (since two have already been picked)
Therefore, the total number of possible combinations is 9 * 8 * 7 = 504.
The probability of the three balls being different colors is 504/504 = 1.
C) To find the probability of Jack winning the game in either the third or fourth attempt, we need to find the probability of winning on the third attempt and the probability of winning on the fourth attempt separately, and then add them together.
The probability of winning on the third attempt is (1/3) * (2/3)^2 = 2/27.
(The first two attempts must be losses, which is (2/3)^2, followed by a win with probability 1/3).
The probability of winning on the fourth attempt is (2/3)^3 * (1/3) = 2/81.
(The first three attempts must be losses, which is (2/3)^3, followed by a win with probability 1/3).
Therefore, the total probability of winning in either the third or fourth attempt is 2/27 + 2/81 = 8/81.