A box contain 12 balls of which some red in ccolour .if 6 more red balls are put in the box and a ball is drawn at random the probability of drwaing aared balls doubles thsn what it was before it was befire find the number of red balls in the bag
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Answered by
7
Hope this helps u.....
Let the total number of red balls initially be x
Total number of balls in the box = 12P(getting a red ball) = x/12
After adding 6 more red balls, the total no:of balls = 12 + 6 = 18 ballsNow, the total number of red balls = (x + 6)
and,P(getting a red ball) = (x + 6)/18
According to the question.
2(x/12) = (x + 6)/18
2x/12 = (x + 6)/18⇒ 36x = 12x + 72
⇒ 36x - 12x = 72
⇒ 24x = 72
x = 3
Let the total number of red balls initially be x
Total number of balls in the box = 12P(getting a red ball) = x/12
After adding 6 more red balls, the total no:of balls = 12 + 6 = 18 ballsNow, the total number of red balls = (x + 6)
and,P(getting a red ball) = (x + 6)/18
According to the question.
2(x/12) = (x + 6)/18
2x/12 = (x + 6)/18⇒ 36x = 12x + 72
⇒ 36x - 12x = 72
⇒ 24x = 72
x = 3
aishiva0204:
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Answered by
15
Hey there !
Let the number of red balls = x.
n(R ) = x.
Probability of red balls = x/12
P ( R) = x/12 .
If 6 red balls are added, then n(R) = x + 6 .
Now, P'(R) = x+6/18 .
x+6/18 =2( x/12)
x + 6 /18 = x/6 .
x + 6 / 3 = x
x + 6 = 3x .
3x - x = 6 .
2x = 6 .
x = 3 .
Number of red balls = 3 .
Hope helped!
Let the number of red balls = x.
n(R ) = x.
Probability of red balls = x/12
P ( R) = x/12 .
If 6 red balls are added, then n(R) = x + 6 .
Now, P'(R) = x+6/18 .
x+6/18 =2( x/12)
x + 6 /18 = x/6 .
x + 6 / 3 = x
x + 6 = 3x .
3x - x = 6 .
2x = 6 .
x = 3 .
Number of red balls = 3 .
Hope helped!
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