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A bag contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. The probability of getting the balls of different colors is:
- a)28/121
- b)56/121
- c)1/2
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A bag contains 7 red and 4 blue balls. Two balls are drawn at random w...
Concept:
- The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as:
- Probability of a Compound Event [(A and B) or (B and C)] is calculated as: P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]('and' means '×' and 'or' means '+')
Calculation:
There are a total of 7 red + 4 blue = 11 balls.
Probability of drawing 1 red ball = 
Probability of drawing 1 blue ball = 
Probability of drawing (1 red) AND (1 blue) ball = 
Similarly, Probability of drawing (1 blue) AND (1 red) ball = 
Probability of getting the balls of different colors = 
Most Upvoted Answer
A bag contains 7 red and 4 blue balls. Two balls are drawn at random w...
Probability Calculation:
To solve this problem, we need to calculate the probability of drawing two balls of different colors.
Step 1: Determine the total number of possible outcomes.
In this case, we are drawing two balls from the bag with replacement, which means we put the first ball back into the bag before drawing the second one.
Since we have 7 red balls and 4 blue balls, the total number of possible outcomes is:
Total outcomes = (Number of balls)^2 = (7 + 4)^2 = 11^2 = 121
Step 2: Determine the number of favorable outcomes.
To calculate the number of favorable outcomes, we need to consider two cases:
1. Drawing a red ball first and then a blue ball.
2. Drawing a blue ball first and then a red ball.
Case 1: Drawing a red ball first and then a blue ball.
The probability of drawing a red ball on the first draw is 7/11 (since there are 7 red balls out of 11 total balls).
After replacing the red ball, the probability of drawing a blue ball on the second draw is also 4/11 (since there are 4 blue balls out of 11 total balls).
Therefore, the probability of drawing a red ball first and then a blue ball is:
P(red then blue) = (7/11) * (4/11) = 28/121
Case 2: Drawing a blue ball first and then a red ball.
The probability of drawing a blue ball on the first draw is 4/11.
After replacing the blue ball, the probability of drawing a red ball on the second draw is 7/11.
Therefore, the probability of drawing a blue ball first and then a red ball is:
P(blue then red) = (4/11) * (7/11) = 28/121
Step 3: Add the probabilities of the two cases.
To find the probability of drawing balls of different colors, we need to add the probabilities of the two cases:
P(different colors) = P(red then blue) + P(blue then red)
= 28/121 + 28/121
= 56/121
Therefore, the probability of getting balls of different colors is 56/121. Hence, the correct answer is option B.
To solve this problem, we need to calculate the probability of drawing two balls of different colors.
Step 1: Determine the total number of possible outcomes.
In this case, we are drawing two balls from the bag with replacement, which means we put the first ball back into the bag before drawing the second one.
Since we have 7 red balls and 4 blue balls, the total number of possible outcomes is:
Total outcomes = (Number of balls)^2 = (7 + 4)^2 = 11^2 = 121
Step 2: Determine the number of favorable outcomes.
To calculate the number of favorable outcomes, we need to consider two cases:
1. Drawing a red ball first and then a blue ball.
2. Drawing a blue ball first and then a red ball.
Case 1: Drawing a red ball first and then a blue ball.
The probability of drawing a red ball on the first draw is 7/11 (since there are 7 red balls out of 11 total balls).
After replacing the red ball, the probability of drawing a blue ball on the second draw is also 4/11 (since there are 4 blue balls out of 11 total balls).
Therefore, the probability of drawing a red ball first and then a blue ball is:
P(red then blue) = (7/11) * (4/11) = 28/121
Case 2: Drawing a blue ball first and then a red ball.
The probability of drawing a blue ball on the first draw is 4/11.
After replacing the blue ball, the probability of drawing a red ball on the second draw is 7/11.
Therefore, the probability of drawing a blue ball first and then a red ball is:
P(blue then red) = (4/11) * (7/11) = 28/121
Step 3: Add the probabilities of the two cases.
To find the probability of drawing balls of different colors, we need to add the probabilities of the two cases:
P(different colors) = P(red then blue) + P(blue then red)
= 28/121 + 28/121
= 56/121
Therefore, the probability of getting balls of different colors is 56/121. Hence, the correct answer is option B.
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Community Answer
A bag contains 7 red and 4 blue balls. Two balls are drawn at random w...
Concept:
- The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as:
- Probability of a Compound Event [(A and B) or (B and C)] is calculated as: P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]('and' means '×' and 'or' means '+')
Calculation:
There are a total of 7 red + 4 blue = 11 balls.
Probability of drawing 1 red ball =
Probability of drawing 1 blue ball =
Probability of drawing (1 red) AND (1 blue) ball =
Similarly, Probability of drawing (1 blue) AND (1 red) ball =
Probability of getting the balls of different colors =
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A bag contains 7 red and 4 blue balls. Two balls are drawn at random withreplacement. The probability of getting the balls of different colors is:a)28/121b)56/121c)1/2d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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