1) A bag contains different colored balls; 7 yellow, 3 red, and 6 green. I randomly select one ball, note its color, put it back in the bag, and select another ball. What is the probability that both balls are green, P(GG)?
A).775
B).15
C).75
D).1406
2) A bag contains different colored balls; 7 yellow, 3 red, and 6 green. I randomly select one ball, note its color, put it back in the bag, and select another ball. What is the probability that both balls are yellow, P(YY)?
A).875
B).2042
C).9042
D).1914
3) A bag contains different colored balls; 7 yellow, 3 red, and 6 green. I randomly select one ball, note its color, leave it out of the bag, and select another ball. What is the probability that the first ball is red and the second is yellow, P(RY)?
A).082
B).0857
C).0375
D).6542
4) A bag contains different colored balls; 7 yellow, 3 red, and 6 green. I randomly select one ball, note its color, leave it out of the bag, and select another ball. What is the probability that the first ball is red and the second is green, P(RG)?
A).0703
B).075
C).5875
D).0857
#1. 16 balls, 6 green, so P(GG) = 6/16 * 6/16 = 9/64 = 0.1406
what do you get on the others?
1) The probability of selecting a green ball on the first draw is 6/16, since there are 6 green balls out of a total of 16 balls. After putting the first ball back in the bag, the probability of selecting another green ball on the second draw is also 6/16. The probability of both balls being green is the product of these two probabilities, so P(GG) = (6/16) * (6/16) = 36/256 = 0.1406. Therefore, the answer is (D) 0.1406.
2) The probability of selecting a yellow ball on the first draw is 7/16, since there are 7 yellow balls out of a total of 16 balls. After putting the first ball back in the bag, the probability of selecting another yellow ball on the second draw is also 7/16. The probability of both balls being yellow is the product of these two probabilities, so P(YY) = (7/16) * (7/16) = 49/256 ≈ 0.1914. Therefore, the answer is (D) 0.1914.
3) The probability of selecting a red ball on the first draw is 3/16, since there are 3 red balls out of a total of 16 balls. After leaving the first ball out of the bag, there are now 15 balls left, with 7 of them being yellow. Therefore, the probability of selecting a yellow ball on the second draw is 7/15. The probability of the first ball being red and the second ball being yellow is the product of these two probabilities, so P(RY) = (3/16) * (7/15) ≈ 0.082. Therefore, the answer is (A) 0.082.
4) The probability of selecting a red ball on the first draw is 3/16, since there are 3 red balls out of a total of 16 balls. After leaving the first ball out of the bag, there are now 15 balls left, with 6 of them being green. Therefore, the probability of selecting a green ball on the second draw is 6/15. The probability of the first ball being red and the second ball being green is the product of these two probabilities, so P(RG) = (3/16) * (6/15) ≈ 0.0703. Therefore, the answer is (A) 0.0703.
To answer these questions, we need to use the concept of probability. The probability of an event happening is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
Let's go through each question step by step.
1) Probability of selecting a green ball on the first draw: P(G1) = 6/(7+3+6) = 6/16 = 3/8
Probability of selecting a green ball on the second draw: P(G2) = 6/16
Probability of both balls being green: P(GG) = P(G1) * P(G2) = (3/8) * (6/16) = 18/128 = 9/64
Answer: The probability that both balls are green is 9/64, which is not one of the given options.
2) Probability of selecting a yellow ball on the first draw: P(Y1) = 7/16
Probability of selecting a yellow ball on the second draw: P(Y2) = 7/16
Probability of both balls being yellow: P(YY) = P(Y1) * P(Y2) = (7/16) * (7/16) = 49/256
Answer: The probability that both balls are yellow is 49/256, which is not one of the given options.
3) Probability of selecting a red ball on the first draw: P(R1) = 3/16
Probability of selecting a yellow ball on the second draw: P(Y2) = 7/15 (since the first ball is left out)
Probability of the first ball being red and the second being yellow: P(RY) = P(R1) * P(Y2) = (3/16) * (7/15) = 21/240 = 7/80
Answer: The probability that the first ball is red and the second is yellow is 7/80, which is not one of the given options.
4) Probability of selecting a red ball on the first draw: P(R1) = 3/16
Probability of selecting a green ball on the second draw: P(G2) = 6/15 (since the red ball is left out)
Probability of the first ball being red and the second being green: P(RG) = P(R1) * P(G2) = (3/16) * (6/15) = 18/240 = 3/40
Answer: The probability that the first ball is red and the second is green is 3/40, which is not one of the given options.
Unfortunately, none of the provided options match the correct probabilities for any of the given scenarios.
To solve these probability problems, we need to understand the concept of probability and use the formula for calculating the probability of independent events.
Probability is the measure of the likelihood that a certain event will occur. It is usually expressed as a number between 0 and 1, where 0 represents an impossible event, and 1 represents a certain event.
The formula for calculating the probability of independent events is:
P(A and B) = P(A) x P(B)
where P(A and B) represents the probability of both event A and event B happening, P(A) represents the probability of event A happening, and P(B) represents the probability of event B happening.
Now, let's solve the given probability problems step by step:
1) Probability of selecting a green ball on the first draw: P(Green) = Number of green balls / Total number of balls = 6 / (7 + 3 + 6) = 6 / 16 = 3/8.
Probability of selecting a green ball on the second draw (note that the first ball is replaced before drawing the second): P(Green) = 6 / 16 = 3/8.
Since the two draws are independent events, the probability of both balls being green is:
P(GG) = P(Green on the first draw) x P(Green on the second draw) = (3/8) x (3/8) = 9/64.
The answer to question 1) is not given as an option, so there might be a mistake in the provided choices.
2) Probability of selecting a yellow ball on the first draw: P(Yellow) = 7 / 16.
Probability of selecting a yellow ball on the second draw (note that the first ball is replaced before drawing the second): P(Yellow) = 7 / 16.
Since the two draws are independent events, the probability of both balls being yellow is:
P(YY) = P(Yellow on the first draw) x P(Yellow on the second draw) = (7/16) x (7/16) = 49/256.
The answer to question 2) is not given as an option, so there might be a mistake in the provided choices.
3) Probability of selecting a red ball on the first draw: P(Red) = 3 / 16.
Probability of selecting a yellow ball on the second draw (note that the first ball is not replaced before drawing the second): P(Yellow) = 7 / 15 (since one red ball is out of the bag).
The probability of the first ball being red and the second ball being yellow is:
P(RY) = P(Red on the first draw) x P(Yellow on the second draw) = (3/16) x (7/15) = 21/320.
The answer to question 3) is not given as an option, so there might be a mistake in the provided choices.
4) Probability of selecting a red ball on the first draw: P(Red) = 3 / 16.
Probability of selecting a green ball on the second draw (note that the first ball is not replaced before drawing the second): P(Green) = 6 / 15 (since one red ball is out of the bag).
The probability of the first ball being red and the second ball being green is:
P(RG) = P(Red on the first draw) x P(Green on the second draw) = (3/16) x (6/15) = 18/240 = 3/40.
The correct answer to question 4) is not provided in the answer choices.
In all of the given questions, there seems to be an inconsistency in the provided answer choices.