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Number of green balls = 12 .
n(G ) = 12 .
Number of Black balls = x.
n(B ) = x.
Total number of balls = 12 + x.
Now, P(B) = x /( 12 + x)
Given that 2 Red balls are added to the bag removing 2 black ones.
n(R ) = 2 .
n(B) = x - 2
Now, P' ( B) = x - 2 /(12 + x) .
Given that,
x-2/12+x = 2/3 ( x / 12 + x)
=> x - 2 = 2/3 ( x)
=> x - 2/3x = 2
=> 1/3x = 2
=> x = 6 .
Therefore value of x = 6 .
n(G ) = 12 .
Number of Black balls = x.
n(B ) = x.
Total number of balls = 12 + x.
Now, P(B) = x /( 12 + x)
Given that 2 Red balls are added to the bag removing 2 black ones.
n(R ) = 2 .
n(B) = x - 2
Now, P' ( B) = x - 2 /(12 + x) .
Given that,
x-2/12+x = 2/3 ( x / 12 + x)
=> x - 2 = 2/3 ( x)
=> x - 2/3x = 2
=> 1/3x = 2
=> x = 6 .
Therefore value of x = 6 .
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