A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag.

Solution

Number of red balls in the bag = 5 Let number of blue balls in the bag = x Total number of balls in the bag = x + 5 i.e., n(S) = x + 5
Let A be the favourable outcomes of getting red balls, then
n(A) = 5
Therefore, P(A) = $fraction numerator straight n left parenthesis straight A right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals fraction numerator 5 over denominator x plus 5 end fraction$

Let B be the favourable outcomes of getting blue balls, then
n(B) = x
Therefore, P(B) = $fraction numerator straight n left parenthesis straight B right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals fraction numerator x over denominator x plus 5 end fraction$
According to equation, P(B) = 2 P(A)
$rightwards double arrow$ $fraction numerator straight x over denominator straight x plus 5 end fraction equals 2 cross times fraction numerator 5 over denominator x plus 5 end fraction rightwards double arrow space space fraction numerator x over denominator x plus 5 end fraction equals open parentheses fraction numerator 10 over denominator x plus 5 end fraction close parentheses rightwards double arrow space space space space x space equals space 10$
Hence, the number of blue balls in the bag is 10.

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Real Numbers

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag.

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