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A bag consists of some red marbles, some blue marbles and two yellow marbles. The probability that a marble picked at random is red is 1 / 2. Also, the probability that two marbles picked simultaneously are blue is 1 / 11. How many marbles are present in the bag?
- a)10
- b)12
- c)15
- d)8
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A bag consists of some red marbles, some blue marbles and two yellow m...
Let the number of marbles in the bag = 2n
Number of yellow marbles = 2
As the probability of drawing a red marble = 1 / 2, number of red marbles in the bag
= n.
Now, number of blue marbles in the bag = 2n - n - 2 = n - 2
Thus, probability of drawing 2 blue marbles
► = n - 2C2 / 2nC2 = ((n - 2)(n - 3))/(2n(2n - 1)) = 1/11
► Or, 11(n - 2)(n - 3) = 2n(2n - 1)
► Or, 7n2 - 53n + 66 = 0
► Or, 7n2 - 42n - 11n + 66 = 0
► Or, 7n(n - 6) - 11(n - 6) = 0
► Or, (7n - 11)(n - 6) = 0 Or, n = 6 (as n is integral)
Thus, number of marbles in the bag = 2n = 2 x 6 = 12.
Number of yellow marbles = 2
As the probability of drawing a red marble = 1 / 2, number of red marbles in the bag
= n.
Now, number of blue marbles in the bag = 2n - n - 2 = n - 2
Thus, probability of drawing 2 blue marbles
► = n - 2C2 / 2nC2 = ((n - 2)(n - 3))/(2n(2n - 1)) = 1/11
► Or, 11(n - 2)(n - 3) = 2n(2n - 1)
► Or, 7n2 - 53n + 66 = 0
► Or, 7n2 - 42n - 11n + 66 = 0
► Or, 7n(n - 6) - 11(n - 6) = 0
► Or, (7n - 11)(n - 6) = 0 Or, n = 6 (as n is integral)
Thus, number of marbles in the bag = 2n = 2 x 6 = 12.
Most Upvoted Answer
A bag consists of some red marbles, some blue marbles and two yellow m...
Let the number of marbles in the bag = 2n
Number of yellow marbles = 2
As the probability of drawing a red marble = 1 / 2, number of red marbles in the bag
= n.
Now, number of blue marbles in the bag = 2n - n - 2 = n - 2
Thus, probability of drawing 2 blue marbles
► = n - 2C2 / 2nC2 = ((n - 2)(n - 3))/(2n(2n - 1)) = 1/11
► Or, 11(n - 2)(n - 3) = 2n(2n - 1)
► Or, 7n2 - 53n + 66 = 0
► Or, 7n2 - 42n - 11n + 66 = 0
► Or, 7n(n - 6) - 11(n - 6) = 0
► Or, (7n - 11)(n - 6) = 0 Or, n = 6 (as n is integral)
Thus, number of marbles in the bag = 2n = 2 x 6 = 12.
Number of yellow marbles = 2
As the probability of drawing a red marble = 1 / 2, number of red marbles in the bag
= n.
Now, number of blue marbles in the bag = 2n - n - 2 = n - 2
Thus, probability of drawing 2 blue marbles
► = n - 2C2 / 2nC2 = ((n - 2)(n - 3))/(2n(2n - 1)) = 1/11
► Or, 11(n - 2)(n - 3) = 2n(2n - 1)
► Or, 7n2 - 53n + 66 = 0
► Or, 7n2 - 42n - 11n + 66 = 0
► Or, 7n(n - 6) - 11(n - 6) = 0
► Or, (7n - 11)(n - 6) = 0 Or, n = 6 (as n is integral)
Thus, number of marbles in the bag = 2n = 2 x 6 = 12.
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Community Answer
A bag consists of some red marbles, some blue marbles and two yellow m...
Let's assume that there are 'x' red marbles, 'y' blue marbles, and 2 yellow marbles in the bag.
Probability of picking a red marble = Number of red marbles / Total number of marbles
So, the probability of picking a red marble is given as 1/2.
=> x / (x + y + 2) = 1/2
Probability of picking two blue marbles simultaneously = (Number of blue marbles / Total number of marbles) * ((Number of blue marbles - 1) / (Total number of marbles - 1))
So, the probability of picking two blue marbles simultaneously is given as 1/11.
=> (y / (x + y + 2)) * ((y - 1) / (x + y + 1)) = 1/11
Solving these two equations, we can find the values of x and y.
Calculation:
From the first equation: x / (x + y + 2) = 1/2
Cross-multiplying, we get:
2x = x + y + 2
x = y + 2
Substituting the value of x in the second equation: (y / (y + 2 + y + 2)) * ((y - 1) / (y + 2 + y + 1)) = 1/11
Simplifying, we get:
(2y - 1) / (4y + 3) = 1/11
11(2y - 1) = 4y + 3
22y - 11 = 4y + 3
18y = 14
y = 14/18
y = 7/9
Since the number of marbles cannot be a fraction, we can say that there are 7 blue marbles in the bag.
Substituting this value in x = y + 2, we get:
x = 7 + 2
x = 9
So, there are 9 red marbles, 7 blue marbles, and 2 yellow marbles in the bag.
The total number of marbles in the bag is:
9 + 7 + 2 = 18
Therefore, the correct answer is option 'B' (12 marbles).
Probability of picking a red marble = Number of red marbles / Total number of marbles
So, the probability of picking a red marble is given as 1/2.
=> x / (x + y + 2) = 1/2
Probability of picking two blue marbles simultaneously = (Number of blue marbles / Total number of marbles) * ((Number of blue marbles - 1) / (Total number of marbles - 1))
So, the probability of picking two blue marbles simultaneously is given as 1/11.
=> (y / (x + y + 2)) * ((y - 1) / (x + y + 1)) = 1/11
Solving these two equations, we can find the values of x and y.
Calculation:
From the first equation: x / (x + y + 2) = 1/2
Cross-multiplying, we get:
2x = x + y + 2
x = y + 2
Substituting the value of x in the second equation: (y / (y + 2 + y + 2)) * ((y - 1) / (y + 2 + y + 1)) = 1/11
Simplifying, we get:
(2y - 1) / (4y + 3) = 1/11
11(2y - 1) = 4y + 3
22y - 11 = 4y + 3
18y = 14
y = 14/18
y = 7/9
Since the number of marbles cannot be a fraction, we can say that there are 7 blue marbles in the bag.
Substituting this value in x = y + 2, we get:
x = 7 + 2
x = 9
So, there are 9 red marbles, 7 blue marbles, and 2 yellow marbles in the bag.
The total number of marbles in the bag is:
9 + 7 + 2 = 18
Therefore, the correct answer is option 'B' (12 marbles).
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A bag consists of some red marbles, some blue marbles and two yellow marbles. The probability that a marble picked at random is red is 1 / 2. Also, the probability that two marbles picked simultaneously are blue is 1 / 11. How many marbles are present in the bag?a)10b)12c)15d)8Correct answer is option 'B'. Can you explain this answer?
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